# Full text of "Wiley Advanced Engineering Mathematics 10th Ed. 2011"

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```Tenth Edition

ERWIN KREYSZIG

MATHEMATICS

Systems of Units. Some Important Conversion Factors

The most important systems of units are shown in the table below. The mks system is also known as
the International System of Units (abbreviated 57), and the abbreviations sec (instead of s),
gm (instead of g), and nt (instead of N) are also used.

System of units

Length

Mass

Time

Force

cgs system

centimeter (cm)

gram (g)

second (s)

dyne

mks system

meter (m)

kilogram (kg)

second (s)

newton (nt)

Engineering system

foot (ft)

slug

second (s)

pound (lb)

1 inch (in.) = 2.540000 cm 1 foot (ft) = 12 in. = 30.480000 cm

1 yard (yd) = 3 ft = 91.440000 cm 1 statute mile (mi) = 5280 ft = 1.609344 km

1 nautical mile = 6080 ft = 1.853184 km

1 acre = 4840 yd 2 = 4046.8564 m 2 1 mi 2 = 640 acres = 2.5899881 km 2

1 fluid ounce = 1/128 U.S. gallon = 231/128 in. 3 = 29.573730 cm 3
1 U.S. gallon = 4 quarts (liq) = 8 pints (liq) = 128 fl oz = 3785.4118 cm 3
1 British Imperial and Canadian gallon = 1.200949 U.S. gallons = 4546.087 cm 3
1 slug = 14.59390 kg

1 pound (lb) = 4.448444 nt 1 newton (nt) = 10 5 dynes

1 British thermal unit (Btu) = 1054.35 joules 1 joule = 10 7 ergs

1 calorie (cal) = 4.1840 joules

1 kilowatt-hour (kWh) = 3414.4 Btu = 3.6 ■ 10 6 joules
1 horsepower (hp) = 2542.48 Btu/h = 178.298 cal/sec = 0.74570 kW
1 kilowatt (kW) = 1000 watts = 3414.43 Btu/h = 238.662 cal/s

°F = °C • 1.8 + 32 1° = 60' = 3600" = 0.017453293 radian

For further details see, for example, D. Halliday, R. Resnick, and J. Walker, Fundamentals of Physics. 9th ed., Hoboken,
N. J: Wiley, 2011. See also AN American National Standard, ASTM/IEEE Standard Metric Practice, Institute of Electrical and
Electronics Engineers, Inc. (IEEE), 445 Hoes Lane, Piscataway, N. J. 08854, website at www.ieee.org.

Differentiation

integration

(cu)' = cu' (c constant)

J uv' dx = uv — J u'v dx (by parts)

(m + v)' = u + v'

r x n+1

I x n dx = + c (n ¥= 1)

J n + 1

(uv)' = u'v + uv'

f — dx = In Ixl + c
J x 11

( u\ U V — uv

U/ ~ y2

f e ax dx = - e ax + c

J a

du du dy

— = — ■ — (Chain rule)

dx dy dx

J sin x dx = —cos x + c
J cos x dx = sin x + c

i— 1
1

II

J tan x dx = —In |cosx| + c
J cot x dx = In |sin x| + c

a

II

J sec x dx = In |sec x + tan x| + c

(e ax ) = ae ax

esc x dx = In |csc x — cot x| + c

( a x )' = a x In a

r dx 1 x

(sin x)' = cos x

2 2 — arctan + c

J x + a a a

(cosx/ = — sinx

r dx x

r-n 2 - arcsin + c

J Vfl 2 - X 2 a

(tanx) = sec x

(colx/ = — csc 2 x

r dx x

r -, 5 s — arcsinh + c

J Vx 2 + a 2 a

(smlix/ = coshx

r dx x

r-s K — arccosh + c

J Vx 2 - a 2 a

(coshx) — sinhx

(In x)' = —

X

J sin 2 x dx = \x — \ sin 2x + c
J cos 2 x dx = |x + \ sin 2x + c

/i n t ^°8a e

(log a x) =

X

J tan 2 x dx = tan x — x + c
J cot 2 x dx = —cot x — x + c

(arcsinx)' = , 1

Vl - x 2

^ \n x dx = x\n x — x + c

(arccosx)' = 7===f

Vl - x 2

J e ax sin bx dx

g ax

= „ „ (a sin bx b cos bx) + c

a 2 + b 2

(arctanx)' = ^ ^ 2

J e m cos bx dx

(arccotx)' = - - ^

e ax

= (a cos bx + b sin bx) + c

c 2 + b 2

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ENGINEERING

MATHEMATICS

10

T H EDITION

ENGINEERING

MATHEMATICS

ERWIN KREYSZIG

Professor of Mathematics
Ohio State University
Columbus, Ohio

In collaboration with

HERBERT KREYSZIG

New York, New York

EDWARD J. NORMINTON

Associate Professor of Mathematics
Carleton University
Ottawa, Ontario

WILEY JOHN WILEY & SONS, INC.

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ISBN 978-0-470-45836-5

Printed in the United States of America

10 987654321

PREFACE

Purpose and Structure of the Book

This book provides a comprehensive, thorough, and up-to-date treatment of engineering
mathematics . It is intended to introduce students of engineering, physics, mathematics,
computer science, and related fields to those areas of applied mathematics that are most
relevant for solving practical problems. A course in elementary calculus is the sole
prerequisite . (However, a concise refresher of basic calculus for the student is included
on the inside cover and in Appendix 3.)

The subject matter is arranged into seven parts as follows:

A. Ordinary Differential Equations (ODEs) in Chapters 1-6

B. Linear Algebra. Vector Calculus. See Chapters 7-10

C. Fourier Analysis. Partial Differential Equations (PDEs). See Chapters 11 and 12

D. Complex Analysis in Chapters 13-18

E. Numeric Analysis in Chapters 19-21
Optimization, Graphs in Chapters 22 and 23

G. Probability, Statistics in Chapters 24 and 25.

These are followed by five appendices: 1. References, 2. Answers to Odd-Numbered
5 Table of Functions. This is shown in a block diagram on the next page.

The parts of the book are kept independent. In addition, individual chapters are kept as
independent as possible. (If so needed, any prerequisites — to the level of individual
sections of prior chapters — are clearly stated at the opening of each chapter.) We give the
instructor maximum flexibility in selecting the material and tailoring it to his or her
need. The book has helped to pave the way for the present development of engineering
mathematics. This new edition will prepare the student for the current tasks and the future
by a modern approach to the areas listed above. We provide the material and learning
tools for the students to get a good foundation of engineering mathematics that will help
them in their careers and in further studies.

General Features of the Book Include:

Simplicity of examples to make the book teachable — why choose complicated
examples when simple ones are as instructive or even better?

Independence of parts and blocks of chapters to provide flexibility in tailoring
courses to specific needs.

Self-contained presentation, except for a few clearly marked places where a proof
would exceed the level of the book and a reference is given instead.

Gradual increase in difficulty of material with no jumps or gaps to ensure an
enjoyable teaching and learning experience.

Modern standard notation to help students with other courses, modem books, and
journals in mathematics, engineering, statistics, physics, computer science, and others.

Furthermore, we designed the book to be a single, self-contained, authoritative, and
convenient source for studying and teaching applied mathematics, eliminating the need
for time-consuming searches on the Internet or time-consuming trips to the library to get
a particular reference book.

VII

Preface

viii

PARTS AND CHA PTERS OF THE BOOK

PART A

Chaps. 1-6

Ordinary Differential Equations (ODEs)

Chap

s. 1-4

Basic Material

Chap. 5

Chap. 6

Series Solutions

Laplace Transforms

PART B

Chaps. 7-10

Linear Algebra.

Vector Calculus

Chap. 7

Chap. 9

Matrices,

Vector Differential

Linear Systems

1

Calculus

1

Chap. 8

Chap. 10

Eigenvalue Problems

Vector Integral Calculus

PART C

Chaps. 11-12

Fourier Analysis. Partial Differential
Equations (PDEs)

Chap. 11

Fourier Analysis
Chap. 12

Partial Differential Equations

PART D

Chaps. 13-18
Complex Analysis,
Potential Theory

Chaps. 13-17
Basic Material

Chap. 18

Potential Theory

PART E

Chaps. 19-21
Numeric Analysis

PART F

Chaps. 22-23
Optimization, Graphs

Chap. 19

Chap. 20

Chap. 21

Chap. 22

Chap. 23

Numerics in

Numeric

Numerics for

Linear Programming

Graphs, Optimization

General

Linear Algebra

ODEs and PDEs

PART G

Chaps. 24-25
Probability, Statistics

Chap. 24

Data Analysis. Probability Theory
Chap. 25

Mathematical Statistics

GUIDES AND MANUALS

Maple Computer Guide
Mathematica Computer Guide

Student Solutions Manual
and Study Guide

Instructor’s Manual

Preface

IX

Four Underlying Themes of the Book

The driving force in engineering mathematics is the rapid growth of technology and the
sciences. New areas — often drawing from several disciplines — come into existence.
Electric cars, solar energy, wind energy, green manufacturing, nanotechnology, risk
management, biotechnology, biomedical engineering, computer vision, robotics, space
travel, communication systems, green logistics, transportation systems, financial
engineering, economics, and many other areas are advancing rapidly. What does this mean
for engineering mathematics? The engineer has to take a problem from any diverse area
and be able to model it. This leads to the first of four underlying themes of the book.

1. Modeling is the process in engineering, physics, computer science, biology,
chemistry, environmental science, economics, and other fields whereby a physical situation
or some other observation is translated into a mathematical model. This mathematical
model could be a system of differential equations, such as in population control (Sec. 4.5),
a probabilistic model (Chap. 24), such as in risk management, a linear programming
problem (Secs. 22.2-22.4) in minimizing environmental damage due to pollutants, a
financial problem of valuing a bond leading to an algebraic equation that has to be solved
by Newton’s method (Sec. 19.2), and many others.

The next step is solving the mathematical problem obtained by one of the many
techniques covered in Advanced Engineering Mathematics.

The third step is interpreting the mathematical result in physical or other terms to
see what it means in practice and any implications.

Finally, we may have to make a decision that may be of an industrial nature or
recommend a public policy. For example, the population control model may imply
the policy to stop fishing for 3 years. Or the valuation of the bond may lead to a
recommendation to buy. The variety is endless, but the underlying mathematics is
surprisingly powerful and able to provide advice leading to the achievement of goals
toward the betterment of society, for example, by recommending wise policies
concerning global warming, better allocation of resources in a manufacturing process,
or making statistical decisions (such as in Sec. 25.4 whether a drug is effective in treating
a disease).

While we cannot predict what the future holds, we do know that the student has to
practice modeling by being given problems from many different applications as is done
in this book. We teach modeling from scratch, right in Sec. 1.1, and give many examples
in Sec. 1.3, and continue to reinforce the modeling process throughout the book.

2. Judicious use of powerful software for numerics (listed in the beginning of Part E)
and statistics (Part G) is of growing importance. Projects in engineering and industrial
companies may involve large problems of modeling very complex systems with hundreds
of thousands of equations or even more. They require the use of such software. However,
our policy has always been to leave it up to the instructor to determine the degree of use of
computers, from none or little use to extensive use. More on this below.

3. The beauty of engineering mathematics. Engineering mathematics relies on
relatively few basic concepts and involves powerful unifying principles. We point them
out whenever they are clearly visible, such as in Sec. 4.1 where we “grow” a mixing
problem from one tank to two tanks and a circuit problem from one circuit to two circuits,
thereby also increasing the number of ODEs from one ODE to two ODEs. This is an
example of an attractive mathematical model because the “growth” in the problem is
reflected by an “increase” in ODEs.

X

Preface

4. To clearly identify the conceptual structure of subject matters. For example,
complex analysis (in Part D) is a field that is not monolithic in structure but was formed
by three distinct schools of mathematics. Each gave a different approach, which we clearly
mark. The first approach is solving complex integrals by Cauchy’s integral formula (Chaps.
13 and 14), the second approach is to use the Laurent series and solve complex integrals
by residue integration (Chaps. 15 and 16), and finally we use a geometric approach of
conformal mapping to solve boundary value problems (Chaps. 17 and 18). Learning the
conceptual structure and terminology of the different areas of engineering mathematics is
very important for three reasons:

a. It allows the student to identify a new problem and put it into the right group of
problems. The areas of engineering mathematics are growing but most often retain their
conceptual structure.

b. The student can absorb new information more rapidly by being able to fit it into the
conceptual structure.

c. Knowledge of the conceptual structure and terminology is also important when using
the Internet to search for mathematical information. Since the search proceeds by putting
in key words (i.e., terms) into the search engine, the student has to remember the important
concepts (or be able to look them up in the book) that identify the application and area
of engineering mathematics.

Big Changes in This Edition

Q Problem Sets Changed

The problem sets have been revised and rebalanced with some problem sets having more
problems and some less, reflecting changes in engineering mathematics. There is a greater
emphasis on modeling. Now there are also problems on the discrete Lourier transform
(in Sec. 11.9).

Q Series Solutions of ODEs, Special Functions and Fourier Analysis Reorganized

Chap. 5, on series solutions of ODEs and special functions, has been shortened. Chap. 1 1
on Lourier Analysis now contains Sturm-Liouville problems, orthogonal functions, and
orthogonal eigenfunction expansions (Secs. 1 1.5, 1 1.6), where they fit better conceptually
(rather than in Chap. 5), being extensions of Fourier’s idea of using orthogonal functions.

€> Openings of Parts and Chapters Rewritten As Well As Parts of Sections

In order to give the student a better idea of the structure of the material (see Underlying
Theme 4 above), we have entirely rewritten the openings of parts and chapters.
Furthermore, large parts or individual paragraphs of sections have been rewritten or new
sentences inserted into the text. This should give the students a better intuitive
understanding of the material (see Theme 3 above), let them draw conclusions on their
own, and be able to tackle more advanced material. Overall, we feel that the book has
become more detailed and leisurely written.

f|H Student Solutions Manual and Study Guide Enlarged

Upon the explicit request of the users, the answers provided are more detailed and
complete. More explanations are given on how to learn the material effectively by pointing
out what is most important.

Q More Historical Footnotes, Some Enlarged

Historical footnotes are there to show the student that many people from different countries
working in different professions, such as surveyors, researchers in industry, etc., contributed

Preface

XI

to the field of engineering mathematics. It should encourage the students to be creative in
their own interests and careers and perhaps also to make contributions to engineering
mathematics.

Further Changes and New Features

Parts of Chap. 1 on first-order ODEs are rewritten. More emphasis on modeling, also
new block diagram explaining this concept in Sec. 1.1. Early introduction of Euler’s
method in Sec. 1.2 to familiarize student with basic numerics. More examples of
separable ODEs in Sec. 1.3.

For Chap. 2, on second-order ODEs, note the following changes: For ease of reading,
the first part of Sec. 2.4, which deals with setting up the mass-spring system, has
been rewritten; also some rewriting in Sec. 2.5 on the Euler-Cauchy equation.

Substantially shortened Chap. 5, Series Solutions of ODEs. Special Functions:
combined Secs. 5.1 and 5.2 into one section called “Power Series Method,” shortened
material in Sec. 5.4 Bessel’s Equation (of the first kind), removed Sec. 5.7
(Sturm-Liouville Problems) and Sec. 5.8 (Orthogonal Eigenfunction Expansions) and
moved material into Chap. 1 1 (see “Major Changes” above).

New equivalent definition of basis (Sec. 7.4).

In Sec. 7.9, completely new part on composition of linear transformations with
two new examples. Also, more detailed explanation of the role of axioms, in
connection with the definition of vector space.

New table of orientation (opening of Chap. 8 “Linear Algebra: Matrix Eigenvalue
Problems”) where eigenvalue problems occur in the book. More intuitive explanation
of what an eigenvalue is at the begining of Sec. 8.1.

Better definition of cross product (in vector differential calculus) by properly
identifying the degenerate case (in Sec. 9.3).

Chap. 11 on Fourier Analysis extensively rearranged: Secs. 11.2 and 11.3

combined into one section (Sec. 11.2), old Sec. 11.4 on complex Fourier Series
removed and new Secs. 11.5 (Sturm-Liouville Problems) and 11.6 (Orthogonal
Series) put in (see “Major Changes” above). New problems (new!) in problem set

11.9 on discrete Fourier transform.

New section 12.5 on modeling heat flow from a body in space by setting up the heat
equation. Modeling PDEs is more difficult so we separated the modeling process
from the solving process (in Sec. 12.6).

Introduction to Numerics rewritten for greater clarity and better presentation; new
Example 1 on how to round a number. Sec. 19.3 on interpolation shortened by
removing the less important central difference formula and giving a reference instead.

Large new footnote with historical details in Sec. 22.3, honoring George Dantzig,
the inventor of the simplex method.

Traveling salesman problem now described better as a “difficult” problem, typical
of combinatorial optimization (in Sec. 23.2). More careful explanation on how to
compute the capacity of a cut set in Sec. 23.6 (Flows on Networks).

In Chap. 24, material on data representation and characterization restructured in
terms of five examples and enlarged to include empirical rule on distribution of

XII

Preface

data, outliers, and the score (Sec. 24.1). Furthermore, new example on encription
(Sec. 24.4).

Lists of software for numerics (Part E) and statistics (Part G) updated.

References in Appendix 1 updated to include new editions and some references to
websites.

Use of Computers

The presentation in this book is adaptable to various degrees of use of software,
Computer Algebra Systems (CAS’s), or programmable graphic calculators, ranging
from no use, very little use, medium use, to intensive use of such technology. The choice
of how much computer content the course should have is left up to the instructor, thereby
exhibiting our philosophy of maximum flexibility and adaptability. And, no matter what
the instructor decides, there will be no gaps or jumps in the text or problem set. Some
problems are clearly designed as routine and drill exercises and should be solved by
hand (paper and pencil, or typing on your computer). Other problems require more
thinking and can also be solved without computers. Then there are problems where the
computer can give the student a hand. And finally, the book has CAS projects, CAS
problems and CAS experiments , which do require a computer, and show its power in
solving problems that are difficult or impossible to access otherwise. Here our goal is
to combine intelligent computer use with high-quality mathematics. The computer
invites visualization, experimentation, and independent discovery work. In summary,
the high degree of flexibility of computer use for the book is possible since there are
plenty of problems to choose from and the CAS problems can be omitted if desired.

Note that information on software (what is available and where to order it) is at the
beginning of Part E on Numeric Analysis and Part G on Probability and Statistics. Since
Maple and Mathematica are popular Computer Algebra Systems, there are two computer
guides available that are specifically tailored to Advanced Engineering Mathematics:
E. Kreyszig and E.J. Norminton, Maple Computer Guide, 10th Edition and Mathematica
Computer Guide. 10 th Edition. Their use is completely optional as the text in the book is
written without the guides in mind.

Suggestions for Courses: A Four-Semester Sequence

The material, when taken in sequence, is suitable for four consecutive semester courses,
meeting 3 to 4 hours a week:

1st Semester
2nd Semester
3rd Semester
4th Semester

ODEs (Chaps. 1-5 or 1-6)

Linear Algebra. Vector Analysis (Chaps. 7-10)
Complex Analysis (Chaps. 13-18)

Numeric Methods (Chaps. 19-21)

Suggestions for Independent One-Semester Courses

The book is also suitable for various independent one-semester courses meeting 3 hours
a week. For instance.

Introduction to ODEs (Chaps. 1-2, 21.1)

Laplace Transforms (Chap. 6)

Matrices and Linear Systems (Chaps. 7-8)

Vector Algebra and Calculus (Chaps. 9-10)

Fourier Series and PDEs (Chaps. 11-12, Secs. 21.4—21.7)

Introduction to Complex Analysis (Chaps. 13-17)

Numeric Analysis (Chaps. 19, 21)

Numeric Linear Algebra (Chap. 20)

Optimization (Chaps. 22-23)

Graphs and Combinatorial Optimization (Chap. 23)

Probability and Statistics (Chaps. 24-25)

Acknowledgments

We are indebted to former teachers, colleagues, and students who helped us directly or
indirectly in preparing this book, in particular this new edition. We profited greatly from
discussions with engineers, physicists, mathematicians, computer scientists, and others,
and from their written comments. We would like to mention in particular Professors

Y. A. Antipov, R. Belinski, S. L. Campbell, R. Carr, P. L. Chambre, Isabel F. Cruz,

Z. Davis, D. Dicker, L. D. Drager, D. Ellis, W. Fox, A. Goriely, R. B. Guenther,
J. B. Handley, N. Harbertson, A. Hassen, V. W. Howe, H. Kuhn, K. Millet, J. D. Moore,
W. D. Munroe, A. Nadim, B. S. Ng, J. N. Ong, P. J. Pritchard, W. O. Ray, L. F. Shampine,
H. L. Smith, Roberto Tamassia, A. L. Villone, H. J. Weiss, A. Wilansky, Neil M. Wigley,
and L. Ying; Maria E. and Jorge A. Miranda, JD, all from the United States; Professors
Wayne H. Enright, Francis. L. Lemire, James J. Little, David G. Lowe, Gerry McPhail,
Theodore S. Norvell, and R. Vaillancourt; Jeff Seiler and David Stanley, all from Canada;
and Professor Eugen Eichhorn, Gisela Heckler, Dr. Gunnar Schroeder, and Wiltrud
Stiefenhofer from Europe. Furthermore, we would like to thank Professors John
B. Donaldson, Bruce C. N. Greenwald, Jonathan L. Gross, Morris B. Holbrook, John
R. Render, and Bernd Schmitt; and Nicholaiv Villalobos, all from Columbia University,
New York; as well as Dr. Pearl Chang, Chris Gee, Mike Hale, Joshua Jayasingh, MD,
David Kahr, Mike Lee, R. Richard Royce, Elaine Schattner, MD, Raheel Siddiqui, Robert
Sullivan, MD, Nancy Veit, and Ana M. Kreyszig, JD, all from New York City. We would
also like to gratefully acknowledge the use of facilities at Carleton University, Ottawa,
and Columbia University, New York.

Furthermore we wish to thank John Wiley and Sons, in particular Publisher Laurie
Rosatone, Editor Shannon Corliss, Production Editor Barbara Russiello, Media Editor
Melissa Edwards, Text and Cover Designer Madelyn Lesure, and Photo Editor Sheena
Goldstein for their great care and dedication in preparing this edition. In the same vein,
we would also like to thank Beatrice Ruberto, copy editor and proofreader, WordCo, for
the Index, and Joyce Franzen of PreMedia and those of PreMedia Global who typeset this
edition.

Suggestions of many readers worldwide were evaluated in preparing this edition.
Further comments and suggestions for improving the book will be gratefully received.

KREYSZIG

CONTENTS

PART A

Ordinary Differential Equations (ODEs) 1

CHAPTER 1 First-Order ODEs 2

1.1 Basic Concepts. Modeling 2

1.2 Geometric Meaning of y = f(x, y). Direction Fields, Euler’s Method 9

1.3 Separable ODEs. Modeling 12

1.4 Exact ODEs. Integrating Factors 20

1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 27

1.6 Orthogonal Trajectories. Optional 36

1.7 Existence and Uniqueness of Solutions for Initial Value Problems 38
Chapter 1 Review Questions and Problems 43

Summary of Chapter 1 44

CHAPTER 2 Second-Order Linear ODEs 46

2.1 Homogeneous Linear ODEs of Second Order 46

2.2 Homogeneous Linear ODEs with Constant Coefficients 53

2.3 Differential Operators. Optional 60

2.4 Modeling of Free Oscillations of a Mass-Spring System 62

2.5 Euler-Cauchy Equations 71

2.6 Existence and Uniqueness of Solutions. Wronskian 74

2.7 Nonhomogeneous ODEs 79

2.8 Modeling: Forced Oscillations. Resonance 85

2.9 Modeling: Electric Circuits 93

2.10 Solution by Variation of Parameters 99
Chapter 2 Review Questions and Problems 102
Summary of Chapter 2 103

CHAPTER 3 Higher Order Linear ODEs 105

3.1 Homogeneous Linear ODEs 105

3.2 Homogeneous Linear ODEs with Constant Coefficients 111

3.3 Nonhomogeneous Linear ODEs 116
Chapter 3 Review Questions and Problems 122
Summary of Chapter 3 123

CHAPTER 4 Systems of ODEs. Phase Plane. Qualitative Methods 124

4.0 For Reference: Basics of Matrices and Vectors 124

4.1 Systems of ODEs as Models in Engineering Applications 130

4.2 Basic Theory of Systems of ODEs. Wronskian 137

4.3 Constant-Coefficient Systems. Phase Plane Method 140

4.4 Criteria for Critical Points. Stability 148

4.5 Qualitative Methods for Nonlinear Systems 152

4.6 Nonhomogeneous Linear Systems of ODEs 160
Chapter 4 Review Questions and Problems 164
Summary of Chapter 4 165

CHAPTER 5 Series Solutions of ODEs. Special Functions 167

5.1 Power Series Method 167

5.2 Legendre’s Equation. Legendre Polynomials P n (x ) 175

xv

XVI

Contents

5.3 Extended Power Series Method: Frobenius Method 180

5.4 Bessel’s Equation. Bessel Functions J v (x ) 187

5.5 Bessel Functions of the Y v (x). General Solution 196
Chapter 5 Review Questions and Problems 200
Summary of Chapter 5 201

CHAPTER 6 Laplace Transforms 203

6.1 Laplace Transform. Linearity. First Shifting Theorem (^-Shifting) 204

6.2 Transforms of Derivatives and Integrals. ODEs 211

6.3 Unit Step Function (Heaviside Function).

Second Shifting Theorem (r-Shifting) 217

6.4 Short Impulses. Dirac’s Delta Function. Partial Fractions 225

6.5 Convolution. Integral Equations 232

6.6 Differentiation and Integration of Transforms.

ODEs with Variable Coefficients 238

6.7 Systems of ODEs 242

6.8 Laplace Transform: General Formulas 248

6.9 Table of Laplace Transforms 249
Chapter 6 Review Questions and Problems 251
Summary of Chapter 6 253

PART B Linear Algebra. Vector Calculus 255

CHAPTER 7 Linear Algebra: Matrices, Vectors, Determinants.

Linear Systems 256

7.1 Matrices, Vectors: Addition and Scalar Multiplication 257

7.2 Matrix Multiplication 263

7.3 Linear Systems of Equations. Gauss Elimination 272

7.4 Linear Independence. Rank of a Matrix. Vector Space 282

7.5 Solutions of Linear Systems: Existence, Uniqueness 288

7.6 For Reference: Second- and Third-Order Determinants 291

7.7 Determinants. Cramer’s Rule 293

7.8 Inverse of a Matrix. Gauss-Jordan Elimination 301

7.9 Vector Spaces, Inner Product Spaces. Linear Transformations. Optional 309
Chapter 7 Review Questions and Problems 318

Summary of Chapter 7 320

CHAPTER 8 Linear Algebra: Matrix Eigenvalue Problems 322

8.1 The Matrix Eigenvalue Problem.

Determining Eigenvalues and Eigenvectors 323

8.2 Some Applications of Eigenvalue Problems 329

8.3 Symmetric, Skew-Symmetric, and Orthogonal Matrices 334

8.4 Eigenbases. Diagonalization. Quadratic Forms 339

8.5 Complex Matrices and Forms. Optional 346
Chapter 8 Review Questions and Problems 352
Summary of Chapter 8 353

Contents

XVII

CHAPTER 9 Vector Differential Calculus. Grad, Div, Curl 354

9.1 Vectors in 2-Space and 3-Space 354

9.2 Inner Product (Dot Product) 361

9.3 Vector Product (Cross Product) 368

9.4 Vector and Scalar Functions and Their Fields. Vector Calculus: Derivatives 375

9.5 Curves. Arc Length. Curvature. Torsion 381

9.6 Calculus Review: Functions of Several Variables. Optional 392

9.7 Gradient of a Scalar Field. Directional Derivative 395

9.8 Divergence of a Vector Field 402

9.9 Curl of a Vector Field 406

Chapter 9 Review Questions and Problems 409
Summary of Chapter 9 410

CHAPTER 10 Vector Integral Calculus. Integral Theorems 413

10.1 Line Integrals 413

10.2 Path Independence of Line Integrals 419

10.3 Calculus Review: Double Integrals. Optional 426

10.4 Green’s Theorem in the Plane 433

10.5 Surfaces for Surface Integrals 439

10.6 Surface Integrals 443

10.7 Triple Integrals. Divergence Theorem of Gauss 452

10.8 Further Applications of the Divergence Theorem 458

10.9 Stokes’s Theorem 463

Chapter 10 Review Questions and Problems 469
Summary of Chapter 10 470

PART C Fourier Analysis. Partial Differential Equations (PDEs) 473

CHAPTER 11 Fourier Analysis 474

11.1 Fourier Series 474

11.2 Arbitrary Period. Even and Odd Functions. Half-Range Expansions 483

11.3 Forced Oscillations 492

11.4 Approximation by Trigonometric Polynomials 495

11.5 Sturm-Liouville Problems. Orthogonal Functions 498

11.6 Orthogonal Series. Generalized Fourier Series 504

11.7 Fourier Integral 510

11.8 Fourier Cosine and Sine Transforms 518

11.9 Fourier Transform. Discrete and Fast Fourier Transforms 522

11.10 Tables of Transforms 534

Chapter 1 1 Review Questions and Problems 537
Summary of Chapter 1 1 538

CHAPTER 12 Partial Differential Equations (PDEs) 540

12.1 Basic Concepts of PDEs 540

12.2 Modeling: Vibrating String, Wave Equation 543

12.3 Solution by Separating Variables. Use of Fourier Series 545

12.4 D’Alembert’s Solution of the Wave Equation. Characteristics 553

12.5 Modeling: Heat Flow from a Body in Space. Heat Equation 557

Contents

xviii

12.6 Heat Equation: Solution by Fourier Series.

Steady Two-Dimensional Heat Problems. Dirichlet Problem 558

12.7 Heat Equation: Modeling Very Long Bars.

Solution by Fourier Integrals and Transforms 568

12.8 Modeling: Membrane, Two-Dimensional Wave Equation 575

12.9 Rectangular Membrane. Double Fourier Series 577

12.10 Laplacian in Polar Coordinates. Circular Membrane. Fourier-Bessel Series 585

12.11 Laplace’s Equation in Cylindrical and Spherical Coordinates. Potential 593

12.12 Solution of PDEs by Laplace Transforms 600
Chapter 12 Review Questions and Problems 603
Summary of Chapter 12 604

PART D Complex Analysis 607

CHAPTER 13 Complex Numbers and Functions.

Complex Differentiation 608

13.1 Complex Numbers and Their Geometric Representation 608

13.2 Polar Form of Complex Numbers. Powers and Roots 613

13.3 Derivative. Analytic Function 619

13.4 Cauchy-Riemann Equations. Laplace’s Equation 625

13.5 Exponential Function 630

13.6 Trigonometric and Hyperbolic Functions. Euler’s Formula 633

13.7 Logarithm. General Power. Principal Value 636
Chapter 13 Review Questions and Problems 641
Summary of Chapter 13 641

CHAPTER 14 Complex Integration 643

14.1 Line Integral in the Complex Plane 643

14.2 Cauchy’s Integral Theorem 652

14.3 Cauchy’s Integral Formula 660

14.4 Derivatives of Analytic Functions 664
Chapter 14 Review Questions and Problems 668
Summary of Chapter 14 669

CHAPTER 15 Power Series, Taylor Series 671

15.1 Sequences, Series, Convergence Tests 671

15.2 Power Series 680

15.3 Functions Given by Power Series 685

15.4 Taylor and Maclaurin Series 690

15.5 Uniform Convergence. Optional 698
Chapter 15 Review Questions and Problems 706
Summary of Chapter 15 706

CHAPTER 16 Laurent Series. Residue Integration 708

16.1 Laurent Series 708

16.2 Singularities and Zeros. Infinity 715

16.3 Residue Integration Method 719

16.4 Residue Integration of Real Integrals 725
Chapter 16 Review Questions and Problems 733
Summary of Chapter 16 734

Contents

XIX

CHAPTER 17 Conformal Mapping 736

17.1 Geometry of Analytic Functions: Conformal Mapping 737

17.2 Linear Fractional Transformations (Mobius Transformations) 742

17.3 Special Linear Fractional Transformations 746

17.4 Conformal Mapping by Other Functions 750

17.5 Riemann Surfaces. Optional 754
Chapter 17 Review Questions and Problems 756
Summary of Chapter 17 757

CHAPTER 18 Complex Analysis and Potential Theory 758

18.1 Electrostatic Fields 759

18.2 Use of Conformal Mapping. Modeling 763

18.3 Heat Problems 767

18.4 Fluid Flow 771

18.5 Poisson’s Integral Formula for Potentials 777

18.6 General Properties of Harmonic Functions.

Uniqueness Theorem for the Dirichlet Problem 781

Chapter 18 Review Questions and Problems 785
Summary of Chapter 18 786

PART E Numeric Analysis 787

Software 788

CHAPTER 19 Numerics in General 790

19.1 Introduction 790

19.2 Solution of Equations by Iteration 798

19.3 Interpolation 808

19.4 Spline Interpolation 820

19.5 Numeric Integration and Differentiation 827
Chapter 19 Review Questions and Problems 841
Summary of Chapter 19 842

CHAPTER 20 Numeric Linear Algebra 844

20.1 Linear Systems: Gauss Elimination 844

20.2 Linear Systems: LU-Factorization, Matrix Inversion 852

20.3 Linear Systems: Solution by Iteration 858

20.4 Linear Systems: Ill-Conditioning, Norms 864

20.5 Least Squares Method 872

20.6 Matrix Eigenvalue Problems: Introduction 876

20.7 Inclusion of Matrix Eigenvalues 879

20.8 Power Method for Eigenvalues 885

20.9 Tridiagonalization and QR-Factorization 888
Chapter 20 Review Questions and Problems 896
Summary of Chapter 20 898

CHAPTER 21 Numerics for ODEs and PDEs 900

21. Methods for First-Order ODEs 901

21.2 Multistep Methods 911

21.3 Methods for Systems and Higher Order ODEs 915

XX

Contents

21.4 Methods for Elliptic PDEs 922

21.5 Neumann and Mixed Problems. Irregular Boundary 931

21.6 Methods for Parabolic PDEs 936

21.7 Method for Hyperbolic PDEs 942
Chapter 21 Review Questions and Problems 945
Summary of Chapter 21 946

PART F Optimization, Graphs 949

CHAPTER 22 Unconstrained Optimization. Linear Programming 950

22.1 Basic Concepts. Unconstrained Optimization: Method of Steepest Descent 951

22.2 Linear Programming 954

22.3 Simplex Method 958

22.4 Simplex Method: Difficulties 962
Chapter 22 Review Questions and Problems 968
Summary of Chapter 22 969

CHAPTER 23 Graphs. Combinatorial Optimization 970

23.1 Graphs and Digraphs 970

23.2 Shortest Path Problems. Complexity 975

23.3 Bellman’s Principle. Dijkstra’s Algorithm 980

23.4 Shortest Spanning Trees: Greedy Algorithm 984

23.5 Shortest Spanning Trees: Prim’s Algorithm 988

23.6 Flows in Networks 991

23.7 Maximum Flow: Ford-Fulkerson Algorithm 998

23.8 Bipartite Graphs. Assignment Problems 1001
Chapter 23 Review Questions and Problems 1006
Summary of Chapter 23 1007

PART G Probability, Statistics 1009

Software 1009

CHAPTER 24 Data Analysis. Probability Theory 1011

24.1 Data Representation. Average. Spread 1011

24.2 Experiments, Outcomes, Events 1015

24.3 Probability 1018

24.4 Permutations and Combinations 1024

24.5 Random Variables. Probability Distributions 1029

24.6 Mean and Variance of a Distribution 1035

24.7 Binomial, Poisson, and Hypergeometric Distributions 1039

24.8 Normal Distribution 1045

24.9 Distributions of Several Random Variables 1051
Chapter 24 Review Questions and Problems 1060
Summary of Chapter 24 1060

CHAPTER 25 Mathematical Statistics 1063

25.1 Introduction. Random Sampling 1063

25.2 Point Estimation of Parameters 1065

25.3 Confidence Intervals 1068

Contents

25.4 Testing Hypotheses. Decisions 1077

25.5 Quality Control 1087

25.6 Acceptance Sampling 1092

25.7 Goodness of Fit. ^ 2 -Test 1096

25.8 Nonparametric Tests 1100

25.9 Regression. Fitting Straight Lines. Correlation 1103
Chapter 25 Review Questions and Problems 1111
Summary of Chapter 25 1112

APPENDIX 1 References A1

APPENDIX 2 Answers to Odd-Numbered Problems

APPENDIX 3 Auxiliary Material A63

A3.1 Formulas for Special Functions A63
A3.2 Partial Derivatives A69
A3.3 Sequences and Series A72

A3.4 Grad, Div, Curl, V 2 in Curvilinear Coordinates A74

APPENDIX 5 Tables A97
INDEX 11

PHOTO CREDITS PI

PART A

Ordinary
Differential
Equations (ODEs)

CHAPTER i
CHAPTER 2
CHAPTER 3
CHAPTER 4
CHAPTER 5
CHAPTER 6

First-Order ODEs
Second-Order Linear ODEs
Higher Order Linear ODEs

Systems of ODEs. Phase Plane. Qualitative Methods
Series Solutions of ODEs. Special Functions
Laplace Transforms

Many physical laws and relations can be expressed mathematically in the form of differential
equations. Thus it is natural that this book opens with the study of differential equations and
their solutions. Indeed, many engineering problems appear as differential equations.

The main objectives of Part A are twofold: the study of ordinary differential equations
and their most important methods for solving them and the study of modeling.

Ordinary differential equations (ODEs) are differential equations that depend on a single
variable. The more difficult study of partial differential equations (PDEs), that is,
differential equations that depend on several variables, is covered in Part C.

Modeling is a crucial general process in engineering, physics, computer science, biology,
medicine, environmental science, chemistry, economics, and other fields that translates a
physical situation or some other observations into a “mathematical model.” Numerous
examples from engineering (e.g., mixing problem), physics (e.g., Newton’s law of cooling),
biology (e.g., Gompertz model), chemistry (e.g., radiocarbon dating), environmental science
(e.g., population control), etc. shall be given, whereby this process is explained in detail,
that is, how to set up the problems correctly in terms of differential equations.

For those interested in solving ODEs numerically on the computer, look at Secs. 21.1-21.3
of Chapter 21 of Part F, that is, numeric methods for ODEs. These sections are kept
independent by design of the other sections on numerics. This allows for the study of
numerics for ODEs directly after Chap. 1 or 2.

1

Basic

Physical

System

Mathematical

Model

Mathematical

Solution

Physical

Interpretation

Fig. 1 Modeling,
solving, interpreting

CHAPTER

First-Order ODEs

Chapter 1 begins the study of ordinary differential equations (ODEs) by deriving them from
physical or other problems (modeling), solving them by standard mathematical methods,
and interpreting solutions and their graphs in terms of a given problem. The simplest ODEs
to be discussed are ODEs of the first order because they involve only the first derivative
of the unknown function and no higher derivatives. These unknown functions will usually
be denoted by y(x) or y(t) when the independent variable denotes time t. The chapter ends
with a study of the existence and uniqueness of solutions of ODEs in Sec. 1.7.

Understanding the basics of ODEs requires solving problems by hand (paper and pencil,
or typing on your computer, but first without the aid of a CAS). In doing so, you will
gain an important conceptual understanding and feel for the basic terms, such as ODEs,
direction field, and initial value problem. If you wish, you can use your Computer Algebra
System (CAS) for checking solutions.

COMMENT. Numerics for first-order ODEs can be studied immediately after this
chapter. See Secs. 21.1-21.2, which are independent of other sections on numerics.

Prerequisite: Integral calculus.

Sections that may be omitted in a shorter course: 1.6, 1.7.

References and Answers to Problems: App. 1 Part A, and App. 2.

Concepts. Modeling

If we want to solve an engineering problem (usually of a physical nature), we first
have to formulate the problem as a mathematical expression in terms of variables,
functions, and equations. Such an expression is known as a mathematical model of the
given problem. The process of setting up a model, solving it mathematically, and
interpreting the result in physical or other terms is called mathematical modeling or,
briefly, modeling.

Modeling needs experience, which we shall gain by discussing various examples and

Now many physical concepts, such as velocity and acceleration, are derivatives. Hence
a model is very often an equation containing derivatives of an unknown function. Such
a model is called a differential equation. Of course, we then want to find a solution (a
function that satisfies the equation), explore its properties, graph it, find values of it, and
interpret it in physical terms so that we can understand the behavior of the physical system
in our given problem. However, before we can turn to methods of solution, we must first
define some basic concepts needed throughout this chapter.

2

SEC. 1.1 Basic Concepts. Modeling

3

An ordinary differential equation (ODE) is an equation that contains one or several
derivatives of an unknown function, which we usually call y(x ) (or sometimes y(t) if the
independent variable is time t). The equation may also contain y itself, known functions
of x (or t ), and constants. For example,

(1)

y = cos x

(2)

ft . r\ —2X

y +9 y = e

( 3 )

t tft 3 '2 n

y y - s y = 0

4

CHAP. 1 First-Order ODEs

EXAMPLE 1

are ordinary differential equations (ODEs). Here, as in calculus, y denotes dy/dx,
y" = d 2 y/dx 2 , etc. The term ordinary distinguishes them from partial differential
equations (PDEs), which involve partial derivatives of an unknown function of two
or more variables. For instance, a PDE with unknown function u of two variables x
and y is

d 2 u

dx 2

+

d 2 u

dy 2

= 0 .

PDEs have important engineering applications, but they are more complicated than ODEs;
they will be considered in Chap. 12.

An ODE is said to be of order n if the nth derivative of the unknown function y is the
highest derivative of y in the equation. The concept of order gives a useful classification
into ODEs of first order, second order, and so on. Thus, (1) is of first order, (2) of second
order, and (3) of third order.

In this chapter we shall consider first-order ODEs. Such equations contain only the
first derivative y' and may contain y and any given functions of x. Hence we can write
them as

(4)

F(x,y,y') = 0

or often in the form

/ =f(x,y).

This is called the explicit form, in contrast to the implicit form (4). For instance, the implicit
ODE x~ 3 y' — 4 y 2 = 0 (where x A 0) can be written explicitly as y' = 4x 3 y 2 .

Concept of Solution

A function

y = h(x)

is called a solution of a given ODE (4) on some open interval a < x < b if h(x) is
defined and differentiable throughout the interval and is such that the equation becomes
an identity if y and y are replaced with h and h , respectively. The curve (the graph) of
h is called a solution curve.

Here, open interval a < x < b means that the endpoints a and b are not regarded as
points belonging to the interval. Also, a < x < b includes infinite intervals — oo < x < b,
a < x < °°, — co<^<oo (the real line) as special cases.

Verification of Solution

Verify that y = c/x(c an arbitrary constant) is a solution of the ODE xy = — y for all x A 0. Indeed, differentiate
y = c/x to get y = — c/x . Multiply this by x, obtaining xy’ = — c/x; thus, xy’ = —y, the given ODE.

SEC. 1.1 Basic Concepts. Modeling

5

Solution by Calculus. Solution Curves

The ODE y = dy/dx = cos* can be solved directly by integration on both sides. Indeed, using calculus,
we obtain y = f cos * dx = sin * + c, where c is an arbitrary constant. This is a family of solutions . Each value
of c, for instance, 2.75 or 0 or —8, gives one of these curves. Figure 3 shows some of them, for c = —3, —2,
-1,0, 1,2, 3,4. ■

EXAMPLE 3 (A) Exponential Growth. (B) Exponential Decay

From calculus we know that y = ce°' 2t has the derivative

y = % = 0.2e°' 2t = 0.2y.
dt

Hence y is a solution of y = 0.2y (Fig. 4A). This ODE is of the form y = ky. With positive-constant k it can
model exponential growth, for instance, of colonies of bacteria or populations of animals. It also applies to
humans for small populations in a large country (e.g., the United States in early times) and is then known as

(B) Similarly, y = —0.2 (with a minus on the right) has the solution y = ce~°' 2t , (Fig. 4B) modeling
exponential decay, as, for instance, of a radioactive substance (see Example 5).

y

2.5 -
2.0 -

1.5 - \ \

1.0 -

0.5 ^ " ' -A

0 2 4 6 8 10 12 14 t

Fig. 4B, Solutions of y' = — 0.2y
in Example 3 (exponential decay)

Fig. 4 A. Solutions of y' = 0.2y
in Example 3 (exponential growth)

1 Named after the English pioneer in classic economics, THOMAS ROBERT MALTHUS (1766-1834).

6

CHAP. 1 First-Order ODEs

EXAMPLE 4

We see that each ODE in these examples has a solution that contains an arbitrary
constant c. Such a solution containing an arbitrary constant c is called a general solution
of the ODE.

(We shall see that c is sometimes not completely arbitrary but must be restricted to some
interval to avoid complex expressions in the solution.)

We shall develop methods that will give general solutions uniquely (perhaps except for
notation). Hence we shall say the general solution of a given ODE (instead of a general
solution).

Geometrically, the general solution of an ODE is a family of infinitely many solution
curves, one for each value of the constant c. If we choose a specific c (e.g., c = 6.45 or 0
or —2.01) we obtain what is called a particular solution of the ODE. A particular solution
does not contain any arbitrary constants.

In most cases, general solutions exist, and every solution not containing an arbitrary
constant is obtained as a particular solution by assigning a suitable value to c. Exceptions
to these rules occur but are of minor interest in applications; see Prob. 16 in Problem
Set 1.1.

Initial Value Problem

In most cases the unique solution of a given problem, hence a particular solution, is
obtained from a general solution by an initial condition y(xo) = yo, with given values
xq and yo> that is used to determine a value of the arbitrary constant c. Geometrically
this condition means that the solution curve should pass through the point (xo, yo)
in the xy-plane. An ODE, together with an initial condition, is called an initial value
problem. Thus, if the ODE is explicit, y = /(x, v), the initial value problem is of
the form

(5) y = fix, y), y(x 0 ) = V 0 .

Initial Value Problem

Solve the initial value problem

, dy

y = -T- = 3 v, y{ 0) = 5.7.

ax

Solution. The general solution is ;y(x) = ce 3x \ see Example 3. From this solution and the initial condition
we obtain y(0) = ce° = c = 5.7. Hence the initial value problem has the solution y(x ) = 5.7e . This is a
particular solution.

More on Modeling

The general importance of modeling to the engineer and physicist was emphasized at the
beginning of this section. We shall now consider a basic physical problem that will show
the details of the typical steps of modeling. Step 1 : the transition from the physical situation
(the physical system) to its mathematical formulation (its mathematical model); Step 2:
the solution by a mathematical method; and Step 3: the physical interpretation of the result.
This may be the easiest way to obtain a first idea of the nature and purpose of differential
equations and their applications. Realize at the outset that your computer (your CAS )
may perhaps give you a hand in Step 2, but Steps 1 and 3 are basically your work.

SEC. 1.1 Basic Concepts. Modeling

7

And Step 2 requires a solid knowledge and good understanding of solution methods
available to you — you have to choose the method for your work by hand or by the
computer. Keep this in mind, and always check computer results for errors (which may
arise, for instance, from false inputs).

Given an amount of a radioactive substance, say, 0.5 g (gram), find the amount present at any later time.

Physical Information. Experiments show that at each instant a radioactive substance decomposes — and is thus
decaying in time — proportional to the amount of substance present.

Step 1. Setting up a mathematical model of the physical process. Denote by y(t) the amount of substance still
present at any time t. By the physical law, the time rate of change y ( t ) = dy/dt is proportional to y(r). This
gives the first-order ODE

dy

(6) -=-ky

where the constant k is positive, so that, because of the minus, we do get decay (as in [B] of Example 3).
The value of k is known from experiments for various radioactive substances (e.g., k = 1.4 • 10 -11 sec -1 ,
approximately, for radium gsR 3 )-

Now the given initial amount is 0.5 g, and we can call the corresponding instant t = 0. Then we have the
initial condition y(0) = 0.5. This is the instant at which our observation of the process begins. It motivates
the term initial condition (which, however, is also used when the independent variable is not time or when
we choose a t other than t = 0). Hence the mathematical model of the physical process is the initial value
problem

dy

(7) — = ~ky, y(0) = 0.5.

dt

Step 2. Mathematical solution. As in (B) of Example 3 we conclude that the ODE (6) models exponential decay
and has the general solution (with arbitrary constant c but definite given k)

(8) y(t) = ce~ kt .

We now determine c by using the initial condition. Since y(0) = c from (8), this gives y(0) = c = 0.5. Hence
the particular solution governing our process is (cf. Fig. 5)

(9) y(t) = 0.5e _fct (k > 0).

Always check your result — it may involve human or computer errors! Verify by differentiation (chain rule!)
that your solution (9) satisfies (7) as well as y(0) = 0.5:

f = -0.5 ke~ kt = -k ■ 0.5e~ kt = - ky , y(0) = 0.5e° = 0.5.

dt

Step 3. Interpretation of result. Formula (9) gives the amount of radioactive substance at time t. It starts from
the correct initial amount and decreases with time because k is positive. The limit of y as t — > oo is zero.

y = 0.5e kt , with k = 1.5 as an example)

8

CHAP. 1 First-Order ODEs

PRQB4.ErM=SFT— 1^1

1-8

CALCULUS

Solve the ODE by integration or by remembering a
differentiation formula.

1. y + 2 sin 2ttx = 0

2. y + jre _x2/2 = 0

3- y' = y

4. y = — 1.5y

5. y = 4e~ x cos x

6 n

■ y = ~y

7. y = cosh 5. 13 jc
8. y "' = e~°' 2x

9-15

VERIFICATION. INITIAL VALUE

PROBLEM (IVP)

(a) Verify that y is a solution of the ODE. (b) Determine

from y the particular solution of the IVP. (c) Graph the

solution of the IVP.

9. y' + 4v = 1.4, y = ce~ 4x + 0.35, y(0) = 2

10. y + 5xy = 0, y = ce~ 25x , v(0) = tt

11 . y = y + e x , y = (x + c)e x , y(0) = \

12. yy = 4x, y 2 - 4x 2 = c(y > 0), y(l) = 4

13. y'=y- y 2 , y = 1 _ x , y(0) = 0.25

1 + ce

14. y' tan* = 2y — 8, y = csin 2 jc + 4, y(^Tr) = 0

15. Find two constant solutions of the ODE in Prob. 13 by
inspection.

16. Singular solution. An ODE may sometimes have an
additional solution that cannot be obtained from the
general solution and is then called a singular solution.
The ODE y' 2 — xy' + y = 0 is of this kind. Show
by differentiation and substitution that it has the
general solution y = cx — c 2 and the singular solution
y = x 2 /4. Explain Fig. 6.

Fig. 6

17-20

MODELING, APPLICATIONS

These problems will give you a first impression of modeling.
Many more problems on modeling follow throughout this
chapter.

17. Half-life. The half-life measures exponential decay.
It is the time in which half of the given amount of
radioactive substance will disappear. What is the half-
life of 2 lsRa (in years) in Example 5?

3.6 days.

(a) Given 1 gram, how much will still be present after
1 day?

(b) After 1 year?

19. Free fall. In dropping a stone or an iron ball, air
resistance is practically negligible. Experiments
show that the acceleration of the motion is constant
(equal to g = 9.80 m/sec 2 = 32 ft/sec 2 , called the
acceleration of gravity). Model this as an ODE for
y(t), the distance fallen as a function of time t. If the
motion starts at time t = 0 from rest (i.e., with velocity
v = y' = 0), show that you obtain the familiar law of
free fall

y = Jgt

2

20. Exponential decay. Subsonic flight. The efficiency
of the engines of subsonic airplanes depends on air
pressure and is usually maximum near 35,000 ft.
Find the air pressure y{x) at this height. Physical
information. The rate of change y (x) is proportional
to the pressure. At 18,000 ft it is half its value
yo = >’(0) at sea level. Hint. Remember from calculus
that if y = e kx , then y' = ke kx = ky. Can you see
without calculation that the answer should be close
to y 0 /4?

Particular solutions and singular
solution in Problem 16

SEC. 1.2 Geometric Meaning of y' = f(x, y). Direction Fields, Euler’s Method

9

U Geometric Meaning of y = f(x, y).
Direction Fields, Eulers Method

A first-order ODE

( 1 )

y =f(x,y )

has a simple geometric interpretation. From calculus you know that the derivative y' (x) of

y(x) is the slope of y(jc). Hence a solution curve of (1) that passes through a point (jc 0 , y 0 )
must have, at that point, the slope y (jc 0 ) equal to the value of/ at that point; that is,

Using this fact, we can develop graphic or numeric methods for obtaining approximate
solutions of ODEs (1). This will lead to a better conceptual understanding of an ODE (1).
Moreover, such methods are of practical importance since many ODEs have complicated
solution formulas or no solution formulas at all, whereby numeric methods are needed.

Graphic Method of Direction Fields. Practical Example Illustrated in Fig. 7. We

can show directions of solution curves of a given ODE (1) by drawing short straight-line
segments (lineal elements) in the xy-plane. This gives a direction field (or slope field)
into which you can then fit (approximate) solution curves. This may reveal typical
properties of the whole family of solutions.

Figure 7 shows a direction field for the ODE

obtained by a CAS (Computer Algebra System) and some approximate solution curves
fitted in.

/(*o) = f(x(h >’o)-

( 2 )

y = y + x

\\\\\\\\ \ \-2 \ v

Fig. 7. Direction field of y' = y + x, with three approximate solution
curves passing through (0, 1), (0, 0), (0, —1), respectively

10

CHAP. 1 First-Order ODEs

If you have no CAS, first draw a few level curves f(x, y ) = const of f(x, y), then parallel
lineal elements along each such curve (which is also called an isocline, meaning a curve
of equal inclination), and finally draw approximation curves fit to the lineal elements.

We shall now illustrate how numeric methods work by applying the simplest numeric
method, that is Euler’s method, to an initial value problem involving ODE (2). First we
give a brief description of Euler’s method.

Numeric Method by Euler

Given an ODE (1) and an initial value vGo) = >’o, Euler’s method yields approximate
solution values at equidistant x- values jt 0 , x\ = xq + h, x 2 = x'o + 2 h, • • ■ , namely,

yi = Jo + hf(x o, y 0 ) (Fig. 8)

J 2 = yi + hf(x i,yi), etc.

In general.

y n = y n - 1 + ¥(xn-i,y n -i )

where the step h equals, e.g., 0.1 or 0.2 (as in Table 1.1) or a smaller value for greater
accuracy.

Fig. 8. First Euler step, showing a solution curve, its tangent at (x 0 , y 0 ),
step h and increment hf(x 0 , y 0 ) in the formula for y-|

Table 1.1 shows the computation of n = 5 steps with step h = 0.2 for the ODE (2) and
initial condition y(0) = 0, corresponding to the middle curve in the direction field. We
shall solve the ODE exactly in Sec. 1.5. For the time being, verify that the initial value
problem has the solution y = e x — x — 1. The solution curve and the values in Table 1.1
are shown in Fig. 9. These values are rather inaccurate. The errors y(x n ) — y n are shown
in Table 1.1 as well as in Fig. 9. Decreasing h would improve the values, but would soon
require an impractical amount of computation. Much better methods of a similar nature
will be discussed in Sec. 21.1.

SEC. 1.2 Geometric Meaning of y' = f[x, y). Direction Fields, Euler’s Method

11

Table 1." Euler method fory’ = y + x,y(0) = 0 for
x = 0, , 1.0 with step h — 0.2

n

x n

y n

y(x n )

Error

0

0.0

0.000

0.000

0.000

1

0.2

0.000

0.021

0.021

2

0.4

0.04

0.092

0.052

3

0.6

0.128

0.222

0.094

4

0.8

0.274

0.426

0.152

5

1.0

0.488

0.718

0.230

y

Fig. 9. Euler method: Approximate values in Table 1.1 and solution curve

P^ROBL=E- M= S ^ T ~ l^Z

1-8

DIRECTION FIELDS, SOLUTION CURVES

Graph a direction field (by a CAS or by hand). In the field
graph several solution curves by hand, particularly those
passing through the given points ( x , y).

1.

/

= l +

y 2 ,

(ttt, 1)

2.

yy‘

’ + Ax

= o,

(1,1), (0, 2)

3.

t

y

= l -

v 2 ,

(0, 0), (2, i)

4.

t

y

= 2y-

- y 2 .

(0, 0), (0, 1), (0, 2), (0, 3)

5.

t

y

= X —

i/y.

(1,5)

6.

t

y

= sin 2

y.

(0, -0.4), (0, 1)

7.

t

y

= e v ' x ,

(2, 2), (3, 3)

8.

t

y

= —2xy,

(0, |), (0, 1), (0, 2)

9-10

ACCURACY OF DIRECTION FIELDS

Direction fields are very useful because they can give you
an impression of all solutions without solving the ODE,
which may be difficult or even impossible. To get a feel for
the accuracy of the method, graph a field, sketch solution
curves in it, and compare them with the exact solutions.

9. y = cos ttx

10. y = —5y^ 2 (Sol. Vy + f x = c)

11. Autonomous ODE. This means an ODE not showing
x (the independent variable) explicitly. (The ODEs in
Probs. 6 and 10 are autonomous.) What will the level
curves /(jc, y) = const (also called isoclines = curves

of equal inclination) of an autonomous ODE look like?
Give reason.

12-15

MOTIONS

Model the motion of a body B on a straight line with
velocity as given, y (f) being the distance of B from a point
y = 0 at time t. Graph a direction field of the model (the
ODE). In the field sketch the solution curve satisfying the
given initial condition.

12. Product of velocity times distance constant, equal to 2,
y(0) = 2.

13. Distance = Velocity X Time, y(l) = 1

14. Square of the distance plus square of the velocity equal
to 1 , initial distance 1 / V2

15. Parachutist. Two forces act on a parachutist, the
attraction by the earth mg (m = mass of person plus
equipment, g = 9.8 m/sec 2 the acceleration of gravity)
and the air resistance, assumed to be proportional to the
square of the velocity v(t). Using Newton’s second law
of motion (mass X acceleration = resultant of the forces),
set up a model (an ODE for v(t )). Graph a direction field
(choosing m and the constant of proportionality equal to 1).
Assume that the parachute opens when v = 10 m/sec.
Graph the corresponding solution in the field. What is the
limiting velocity? Would the parachute still be sufficient
if the air resistance were only proportional to u(r)?

12

CHAP. 1 First-Order ODEs

16. CAS PROJECT. Direction Fields. Discuss direction
fields as follows.

(a) Graph portions of the direction field of the ODE (2)
(see Fig. 7), for instance, —5 £ x £ 2, —1 £ y £ 5.
Explain what you have gained by this enlargement of
the portion of the field.

(b) Using implicit differentiation, find an ODE with
the general solution x 2 + 9y 2 = c (y > 0). Graph its
direction field. Does the field give the impression
that the solution curves may be semi-ellipses? Can you
do similar work for circles? Hyperbolas? Parabolas?
Other curves?

(c) Make a conjecture about the solutions of y ' = ~x/y
from the direction field.

(d) Graph the direction field of y = — \y and some
solutions of your choice. How do they behave? Why
do they decrease for y > 0?

17-20

EULER’S METHOD

This is the simplest method to explain numerically solving
an ODE, more precisely, an initial value problem (IVP).
(More accurate methods based on the same principle are
explained in Sec. 21.1.) Using the method, to get a feel for
numerics as well as for the nature of IVPs, solve the IVP
numerically with a PC or a calculator, 10 steps. Graph the
computed values and the solution curve on the same
coordinate axes.

17.

r

y =

: y.

y(0) = l.

h

=

0.1

18.

t

y =

: y.

v(0) = l.

h

=

0.01

19.

/

y =

= O’

1

^”tS3

O

=

0,

h = 0.1

Sol.

y =

- x — tanh x

20.

/

y =

= -

5x 4 y 2 , y(0)

=

1,

h = 0.2

Sol.

y =

= l/d + xf

1.] Separable ODEs. Modeling

Many practically useful ODEs can be reduced to the form

(l) g(y)y' = f(x)

by purely algebraic manipulations. Then we can integrate on both sides with respect to x,
obtaining

( 2 )

g(y)y'dx

f(x) dx + c.

On the left we can switch to y as the variable of integration. By calculus, y dx = dy, so that

( 3 )

g(y)dy

f(x ) dx + c.

If / and g are continuous functions, the integrals in (3) exist, and by evaluating them we
obtain a general solution of (1). This method of solving ODEs is called the method of
separating variables, and (1) is called a separable equation, because in (3) the variables
are now separated: x appears only on the right and y only on the left.

E X A M Separable ODE

The ODE y = 1 + y 2 is separable because it can be written
dy

= dx. By integration, arctan y = x + c or y = tan (x + c ).

1 + y 2

It is very important to introduce the constant of integration immediately when the integration is performed.
If we wrote arctan y = x, then y = tan x, and then introduced c, we would have obtained y = tan x + c, which
is not a solution (when c =£ 0). Verify this.

SEC. 1.3 Separable ODEs. Modeling

13

EXAMPLE 2

EXAMPLE 3

EXAMPLE 4

Separable ODE

The ODE y = (x + 1 )e~ x y 2 is separable; we obtain y~ 2 dy = (x + \)e~ x dx.
By integration, — y -1 = —Or + 2)e~ x + c, y =

1

(x + 2)e — c

Initial Value Problem (IVP). Bell-Shaped Curve

Solve y' = —2xy,y(0) = 1.8.

Solution. By separation and integration.

dy

— = —2 x dx,
y

lny = — x 2 + c, y = ce .

This is the general solution. From it and the initial condition, y(0) = ce u = c = 1.8. Hence the IVP has the
solution y = 1.8e ~ x . This is a particular solution, representing a bell-shaped curve (Fig. 10).

Fig. 10. Solution in Example 3 (bell-shaped curve)

Modeling

The importance of modeling was emphasized in Sec. 1.1, and separable equations yield
various useful models. Let us discuss this in terms of some typical examples.

In September 1991 the famous Iceman (Oetzi), a mummy from the Neolithic period of the Stone Age found in
the ice of the Oetztal Alps (hence the name “Oetzi”) in Southern Tyrolia near the Austrian-Italian border, caused
a scientific sensation. When did Oetzi approximately live and die if the ratio of carbon X gC to carbon 1 §C in
this mummy is 52.5% of that of a living organism?

Physical Information. In the atmosphere and in living organisms, the ratio of radioactive carbon gC (made
radioactive by cosmic rays) to ordinary carbon *§0 is constant. When an organism dies, its absorption of X gC
by breathing and eating terminates. Hence one can estimate the age of a fossil by comparing the radioactive
carbon ratio in the fossil with that in the atmosphere. To do this, one needs to know the half-life of gC, which
is 5715 years ( CRC Handbook of Chemistry and Physics, 83rd ed., Boca Raton: CRC Press, 2002, page 11-52,
line 9).

Solution. Modeling. Radioactive decay is governed by the ODE y' = ky (see Sec. 1.1, Example 5). By
separation and integration (where t is time and yg is the initial ratio of gC to gC)

— = kdt, In \y\ = kt + c, y = yoe kt (yo = <? c )-

y

2 Method by WILLARD FRANK LIBBY (1908-1980), American chemist, who was awarded for this work
the 1960 Nobel Prize in chemistry.

14

CHAP. 1 First-Order ODEs

EXAMPLE 5

Next we use the half-life H = 5715 to determine k. When t = H, half of the original substance is still present. Thus,

J.TT In 0.5 0.693

y 0 e kH = 0.5^o, e kH = 0.5, k = = — = -0.0001213.

H 5715

Finally, we use the ratio 52.5% for determining the time t when Oetzi died (actually, was killed),

e kt = £ -0 - 0001213t = 0.525, t = = 5312. Answer: About 5300 years ago.

-0.0001213 J 6

Other methods show that radiocarbon dating values are usually too small. According to recent research, this is
due to a variation in that carbon ratio because of industrial pollution and other factors, such as nuclear testing.

Mixing Problem

Mixing problems occur quite frequently in chemical industry. We explain here how to solve the basic model
involving a single tank. The tank in Fig. 1 1 contains 1000 gal of water in which initially 100 lb of salt is dissolved.
Brine runs in at a rate of 10 gal/min, and each gallon contains 5 lb of dissoved salt. The mixture in the tank is
kept uniform by stirring. Brine runs out at 10 gal/min. Find the amount of salt in the tank at any time t.

Solution. Step 1. Setting up a model. Let y(f) denote the amount of salt in the tank at time t. Its time rate
of change is

y — Salt inflow rate — Salt outflow rate Balance law.

5 lb times 10 gal gives an inflow of 50 lb of salt. Now, the outflow is 10 gal of brine. This is 10/1000 = 0.01
(= 1%) of the total brine content in the tank, hence 0.01 of the salt content y(t), that is, 0.01 y(t). Thus the
model is the ODE

(4) y = 50 - O.Oly = -0.01(y - 5000).

Step 2. Solution of the model. The ODE (4) is separable. Separation, integration, and taking exponents on both
sides gives

— - = -0.01 dt. In Iv - 5000 1 = -O.Olf + c*, y- 5000 = ce~ omt .

y — 5000

Initially the tank contains 100 lb of salt. Hence y(0) = 100 is the initial condition that will give the unique
solution. Substituting y = 100 and t = 0 in the last equation gives 100 — 5000 = ce = c. Hence c = —4900.
Hence the amount of salt in the tank at time t is

(5) y(t) = 5000 - 4900e~ o olt .

This function shows an exponential approach to the limit 5000 lb; see Fig. 11. Can you explain physically that
y(r) should increase with time? That its limit is 5000 lb? Can you see the limit directly from the ODE?

The model discussed becomes more realistic in problems on pollutants in lakes (see Problem Set 1.5, Prob. 35)
or drugs in organs. These types of problems are more difficult because the mixing may be imperfect and the flow
rates (in and out) may be different and known only very roughly.

Tank

Fig. 11. Mixing problem in Example 5

SEC. 1.3 Separable ODEs. Modeling

15

EXAMPLE 6

Heating an Office Building (Newton’s Law of Cooling 3 )

Suppose that in winter the daytime temperature in a certain office building is maintained at 70°F. The heating
is shut off at 10 P.M. and turned on again at 6 a.m. On a certain day the temperature inside the building at 2 A.M.
was found to be 65 °F. The outside temperature was 50°F at 10 P.M. and had dropped to 40°F by 6 a.m. What
was the temperature inside the building when the heat was turned on at 6 A.M.?

Physical information. Experiments show that the time rate of change of the temperature T of a body B (which
conducts heat well, for example, as a copper ball does) is proportional to the difference between T and the
temperature of the surrounding medium (Newton’s law of cooling).

Solution. Step 1. Setting up a model. Let T{t) be the temperature inside the building and T A the outside
temperature (assumed to be constant in Newton’s law). Then by Newton’s law,

dT ,

(6) — = k(T ~ T a . ).

dt

Such experimental laws are derived under idealized assumptions that rarely hold exactly. However, even if a
model seems to fit the reality only poorly (as in the present case), it may still give valuable qualitative information.
To see how good a model is, the engineer will collect experimental data and compare them with calculations
from the model.

Step 2. General solution. We cannot solve (6) because we do not know T A , just that it varied between 50°F
and 40°F, so we follow the Golden Rule: If you cannot solve your problem, try to solve a simpler one. We
solve (6) with the unknown function T A replaced with the average of the two known values, or 45 °F. For physical
reasons we may expect that this will give us a reasonable approximate value of T in the building at 6 a.m.

For constant T A = 45 (or any other constant value) the ODE (6) is separable. Separation, integration, and
taking exponents gives the general solution

t f = k dt. In | T — 45 1 = kt + c*, T(t) = 45 + ce kt (c = e c ').

Step 3. Particular solution. We choose 10 P.M. to be t = 0. Then the given initial condition is 7(0) = 70 and
yields a particular solution, call it T p . By substitution,

T( 0) = 45 + ce° = 70, c = 70 - 45 = 25, T v (t) = 45 + 25e fct .

Step 4. Determination ofk. We use 7(4) = 65, where t = 4 is 2 a.m. Solving algebraically for k and inserting
k into T V (J) gives (Fig. 12)

T p (A) = 45 + 25e 4fc = 65, e 4fc = 0.8, k = \ In 0.8 = -0.056, T v {t) = 45 + 25e~ 0056t .

Fig. 12. Particular solution (temperature) in Example 6

3 Sir ISAAC NEWTON (1642-1727), great English physicist and mathematician, became a professor at
Cambridge in 1669 and Master of the Mint in 1699. He and the German mathematician and philosopher
GOTTFRIED WILHELM LEIBNIZ (1646-1716) invented (independently) the differential and integral calculus.
Newton discovered many basic physical laws and created the method of investigating physical problems by
means of calculus. His Philosophiae naturalis principia mathematica (. Mathematical Principles of Natural
Philosophy, 1687) contains the development of classical mechanics. His work is of greatest importance to both
mathematics and physics.

16

CHAP. 1 First-Order ODEs

EXAMPLE 7

Step 5. Answer and interpretation. 6 A.M. is r = 8 (namely. 8 hours after 10 P.M.), and

T p ( 8) = 45 + 25e -0 ' 056 ' 8 = 61[°F],

Hence the temperature in the building dropped 9°F, a result that looks reasonable.

Leaking Tank. Outflow of Water Through a Hole (Torricelli’s Law)

This is another prototype engineering problem that leads to an ODE. It concerns the outflow of water from a
cylindrical tank with a hole at the bottom (Fig. 13). You are asked to find the height of the water in the tank at
any time if the tank has diameter 2 m, the hole has diameter 1 cm, and the initial height of the water when the
hole is opened is 2.25 m. When will the tank be empty?

Physical information. Under the influence of gravity the outflowing water has velocity

(7) v(t) = 0.600 \/2 gh(t) (Torricelli’s law 4 ),

where h(t) is the height of the water above the hole at time t, and g = 980cm/sec 2 — 32.17 ft/sec 2 is the
acceleration of gravity at the surface of the earth.

Solution. Step 1. Setting up the model. To get an equation, we relate the decrease in water level h(t ) to the
outflow. The volume AV of the outflow during a short time At is

AV = Av At (A = Area of hole).

AV must equal the change AV* of the volume of the water in the tank. Now

AV* = — B Ah ( B = Cross-sectional area of tank)

where Ah (> 0) is the decrease of the height h{t) of the water. The minus sign appears because the volume of
the water in the tank decreases. Equating AV and AV* gives

-B Ah = Av At.

We now express v according to Torricelli’s law and then let At (the length of the time interval considered)
approach 0 — this is a standard way of obtaining an ODE as a model. That is, we have

Ah A
~At = ~~B V

-f 0.600V2 gh{t)
B

and by letting At — > 0 we obtain the ODE

— = -26.56 -VS.
dt B

where 26.56 = O.6OOV2 ■ 980. This is our model, a first-order ODE.

Step 2. General solution. Our ODE is separable. A/B is constant. Separation and integration gives

dh A A

— — = —26.56 — dt and 2 \fh — c* — 26.56 — t.

Vh B B

Dividing by 2 and squaring gives h = (c — 1 3. 2SAt/B) 2 . Inserting 13.28 A/B = 13.28 • 0.5 2 7r/100 2 7r = 0.000332
yields the general solution

hit) = (c - 0.000 332tf.

4 EVANGELISTA TORRICELLI (1608-1647), Italian physicist, pupil and successor of GALILEO GALILEI
(1564-1642) at Florence. The “contraction factor” 0.600 was introduced by J. C. BORDA in 1766 because the
stream has a smaller cross section than the area of the hole.

SEC. 1.3 Separable ODEs. Modeling

17

Step 3. Particular solution. The initial height (the initial condition) is h{ 0) = 225 cm. Substitution of t = 0
and h = 225 gives from the general solution c 2, = 225, c = 15.00 and thus the particular solution (Fig. 13)

hJt) = (15.00 - 0.0003320

Step 4. Tank empty. h p (t ) = 0 if t = 15.00/0.000332 = 45,181
Here you see distinctly the importance of the choice of units-
in which time is measured in seconds! We used g = 980 cm/sec 2

= 12.6 [hours].

we have been working with the cgs system,

Step 5. Checking. Check the result.

2.25 m

^—2.00

Water
at ti

. i

|

hit

^

Outflowing

water

h

250 -
200 -
150 -
100 -
50 -

0 I I I ! J

0 10000 30000 50000 t

Tank Water level h(t) in tank

Fig. 13 Example 7. Outflow from a cylindrical tank (“leaking tank").
Torricelli's law

Extended Method: Reduction to Separable Form

Certain nonseparable ODEs can be made separable by transformations that introduce for
y a new unknown function. We discuss this technique for a class of ODEs of practical
importance, namely, for equations

( 8 )

Here, /is any (differentiable) function of y/x, such as sin(y/x), (y/x) 4 , and so on. (Such
an ODE is sometimes called a homogeneous ODE, a term we shall not use but reserve
for a more important purpose in Sec. 1.5.)

The form of such an ODE suggests that we set y/x = u\ thus,

(9) y = ux and by product differentiation y' = u x + u.

Substitution into y = f(y/x) then gives u x + u = f{u) or u x = f(u) — u. We see that
if f{u) w f 0, this can be separated:

du dx

f(u) — u x

( 10 )

18

CHAP. 1 First-Order ODEs

EXAMPLE 8 Reduction to Separable Form

Solve

/ 2 2
2xyy = y — x .

Solution. To get the usual explicit form, divide the given equation by 2 xy,

2 2

y -x y x

2 xy 2x 2 y

Now substitute y and y from (9) and then simplify by subtracting u on both sides.

, u 1

U X + u = — — — ,
2 2 u

, u 1 —u — 1
2 2 u 2u

You see that in the last equation you can now separate the variables,
2 u du

1 + u

dx

x

By integration.

In (1 + w 2 ) = -In \x\ + c* = In

+ c*.

Take exponents on both sides to get 1 + u 2 = c/x or 1 + (y/x) 2 = c/x. Multiply the last equation by x 2 to
obtain (Fig. 14)

2 , 2

x + y = cx.

Thus

- y = -

This general solution represents a family of circles passing through the origin with centers on the x-axis.

Fig. 14. General solution (family of circles) in Example 8

PKO&L E : M= S E T ITS

1. CAUTION! Constant of integration. Why is it

important to introduce the constant of integration
immediately when you integrate?

2-10

GENERAL SOLUTION

Find a general solution. Show the steps of derivation. Check

2. y y + x =0

3. y = sec y

4. y sin 27 tx = Try cos 2ttx

5. yy + 36x = 0

6. y = e*-y

7. xy' = y + 2x 3 sin 2 — (Set y/x = u)

8. y = (y + 4x) 2 (Set y + 4x = v)

9. xy = y z + y (Set y/x = u)

10. xy' = x + y (Sety/.r = u)

INITIAL VALUE PROBLEMS (IVPs)

Solve the IVP. Show the steps of derivation, beginning with
the general solution.

11-17

11. xy' + y = 0, y(4) = 6

12. y = 1 + 4y 2 , y(l) = 0

13. y'cosh 2 * = sin 2 y, y(0) = \t t

14. dr/dt = —2 tr, r( 0) = r 0

15. y' = -4 x/y, y(2) = 3

16. y' = (x + y - 2) 2 , y(0) = 2

(Set v = x + y - 2)

17. xy' = y + 3x 4 cos 2 (y/x), y(l) = 0
(Set y/x = u )

18. Particular solution. Introduce limits of integration in
(3) such that y obtained from (3) satisfies the initial
condition y(x 0 ) = yo-

SEC. 1.3 Separable ODEs. Modeling

19

19-36

MODELING, APPLICATIONS

19. Exponential growth. If the growth rate of the number
of bacteria at any time t is proportional to the number
present at t and doubles in 1 week, how many bacteria
can be expected after 2 weeks? After 4 weeks?

20. Another population model.

(a) If the birth rate and death rate of the number of
bacteria are proportional to the number of bacteria
present, what is the population as a function of time.

(b) What is the limiting situation for increasing time?
Interpret it.

21. Radiocarbon dating. What should be the J |C content
(in percent of y 0 ) of a fossilized tree that is claimed to
be 3000 years old? (See Example 4.)

22. Linear accelerators are used in physics for
accelerating charged particles. Suppose that an alpha
particle enters an accelerator and undergoes a constant
acceleration that increases the speed of the particle
from 10 3 m/sec to 10 4 m/sec in 10 -3 sec. Find the
acceleration a and the distance traveled during that
period of 10 -3 sec.

23. Boyle-Mariotte’s law for ideal gases. 5 Experiments
show for a gas at low pressure p (and constant
temperature) the rate of change of the volume V(p)
equals ~V/p. Solve the model.

24. Mixing problem. A tank contains 400 gal of brine
in which 100 lb of salt are dissolved. Fresh water runs
into the tank at a rate of 2 gal/min.The mixture, kept
practically uniform by stirring, runs out at the same
rate. How much salt will there be in the tank at the
end of 1 hour?

25. Newton’s law of cooling. A thermometer, reading
5°C, is brought into a room whose temperature is 22°C.
One minute later the thermometer reading is 12°C.
How long does it take until the reading is practically
22°C, say, 21.9°C?

26. Gompertz growth in tumors. The Gompertz model
is y = —Ay In y (A > 0), where y(t) is the mass of
tumor cells at time t. The model agrees well with
clinical observations. The declining growth rate with
increasing y > 1 corresponds to the fact that cells in
the interior of a tumor may die because of insufficient
oxygen and nutrients. Use the ODE to discuss the
growth and decline of solutions (tumors) and to find
constant solutions. Then solve the ODE.

27. Dryer. If a wet sheet in a dryer loses its moisture at
a rate proportional to its moisture content, and if it
loses half of its moisture during the first 10 min of

drying, when will it be practically dry, say, when will
it have lost 99% of its moisture? First guess, then
calculate.

28. Estimation. Could you see, practically without calcu-
lation, that the answer in Prob. 27 must lie between
60 and 70 min? Explain.

29. Alibi? Jack, arrested when leaving a bar, claims that
he has been inside for at least half an hour (which
would provide him with an alibi). The police check
the water temperature of his car (parked near the
entrance of the bar) at the instant of arrest and again
30 min later, obtaining the values 190°F and 110°F,
respectively. Do these results give Jack an alibi?
(Solve by inspection.)

30. Rocket. A rocket is shot straight up from the earth,
with a net acceleration (= acceleration by the rocket
engine minus gravitational pullback) of 7fm/sec 2
during the initial stage of flight until the engine cut out
at t — 10 sec. How high will it go, air resistance
neglected?

31. Solution curves of y' = g(y/x). Show that any
(nonvertical) straight line through the origin of the
xy-plane intersects all these curves of a given ODE at
the same angle.

32. Friction. If a body slides on a surface, it experiences
friction F (a force against the direction of motion).
Experiments show that |E| = /a|(V| (Coulomb’s 6 law of
kinetic friction without lubrication), where N is the
normal force (force that holds the two surfaces together;
see Fig. 15) and the constant of proportionality p is
called the coefficient of kinetic friction. In Fig. 15
assume that the body weighs 45 nt (about 10 lb; see
front cover for conversion), p = 0.20 (corresponding
to steel on steel), a = 30°, the slide is 10 m long, the
initial velocity is zero, and air resistance is
negligible. Find the velocity of the body at the end
of the slide.

5 R0BERT BOYLE (1627-1691), English physicist and chemist, one of the founders of the Royal Society. EDME MARIOTTE (about
1620-1684), French physicist and prior of a monastry near Dijon. They found the law experimentally in 1662 and 1676, respectively.

e CHARLES AUGUSTIN DE COULOMB (1736-1806). French physicist and engineer.

20

CHAP. 1 First-Order ODEs

33. Rope. To tie a boat in a harbor, how many times
must a rope be wound around a bollard (a vertical
rough cylindrical post fixed on the ground) so that a
man holding one end of the rope can resist a force
exerted by the boat 1000 times greater than the man
can exert? First guess. Experiments show that the
change AS of the force 5 in a small portion of the
rope is proportional to S and to the small angle \(f>
in Fig. 16. Take the proportionality constant 0.15.
The result should surprise you!

34. TEAM PROJECT. Family of Curves. A family of
curves can often be characterized as the general
solution of y = f(x, y).

(a) Show that for the circles with center at the origin
we get y' = —x/y.

(b) Graph some of the hyperbolas xy = c. Find an
ODE for them.

(c) Find an ODE for the straight lines through the
origin.

(d) You will see that the product of the right sides of
the ODEs in (a) and (c) equals — 1 . Do you recognize

this as the condition for the two families to be
orthogonal (i.e., to intersect at right angles)? Do your
graphs confirm this?

(e) Sketch families of curves of your own choice and
find their ODEs. Can every family of curves be given
by an ODE?

35. CAS PROJECT. Graphing Solutions. A CAS can

usually graph solutions, even if they are integrals that
cannot be evaluated by the usual analytical methods of
calculus.

(a) Show this for the five initial value problems
y — e~ x , y(0) = 0, ±1, ±2 graphing all five curves
on the same axes.

(b) Graph approximate solution curves, using the first
few terms of the Maclaurin series (obtained by term-
wise integration of that of y') and compare with the
exact curves.

(c) Repeat the work in (a) for another ODE and initial
that cannot be evaluated as indicated.

36. TEAM PROJECT. Torricelli’s Law. Suppose that
the tank in Example 7 is hemispherical, of radius R,
initially full of water, and has an outlet of 5 cm 2 cross-
sectional area at the bottom. (Make a sketch.) Set
up the model for outflow. Indicate what portion of
your work in Example 7 you can use (so that it can
become part of the general method independent of the
shape of the tank). Find the time t to empty the tank
(a) for any R, (b) for R = 1 m. Plot t as function of
R. Find the time when h = R/2 (a) for any R, (b) for
R = 1 m.

Exact ODEs. Integrating Factors

We recall from calculus that if a function u(x, y ) has continuous partial derivatives, its
differential (also called its total differential') is

du dll

du = — dx H dy.

dx dy

From this it follows that if u(x, y) = c = const, then du = 0.

For example, if u = x + x 2 y 3 = c, then

du = (1 + 2xy 3 ) dx + 3x 2 y 2 dy = 0

or

i dy 1 + 2 xy 3
y dx 3 x 2 y 2

SEC. 1.4 Exact ODEs. Integrating Factors

21

an ODE that we can solve by going backward. This idea leads to a powerful solution
method as follows.

A first-order ODE Mix, y ) + Nix, y)y = 0, written as (use dy = y dx as in Sec. 1.3)

(1) M(x, y) dx + N(x, y) dy = 0

is called an exact differential equation if the differential form Mix, y) dx + N(x, y) dy
is exact, that is, this form is the differential

du du

(2) du = — dx H dy

dx dy

of some function u(x, y). Then (1) can be written

du = 0.

By integration we immediately obtain the general solution of (1) in the form
(3) u(x, y) = c.

This is called an implicit solution, in contrast to a solution y = h(x) as defined in Sec.
1.1, which is also called an explicit solution, for distinction. Sometimes an implicit solution
can be converted to explicit form. (Do this for x z + y 2 = 1.) If this is not possible, your
CAS may graph a figure of the contour lines (3) of the function m(x, y) and help you in
understanding the solution.

Comparing (1) and (2), we see that (1) is an exact differential equation if there is some
function u(x, y) such that

du du

(4) (a) — = M, (b) — = N.

dx dy

From this we can derive a formula for checking whether (1) is exact or not, as follows.

Let M and N be continuous and have continuous first partial derivatives in a region in
the xy- plane whose boundary is a closed curve without self-intersections. Then by partial
differentiation of (4) (see App. 3.2 for notation),

dM _ d 2 u
dy dy dx’

dN d 2 u

dx dx dy

By the assumption of continuity the two second partial derivaties are equal. Thus

dM _ dN
dy dx

( 5 )

22

CHAP. 1 First-Order ODEs

EXAMPLE 1

This condition is not only necessary but also sufficient for (1) to be an exact differential
equation. (We shall prove this in Sec. 10.2 in another context. Some calculus books, for
instance, [GenRef 12], also contain a proof.)

If (1) is exact, the function u(x,y) can be found by inspection or in the following
systematic way. From (4a) we have by integration with respect to x

( 6 )

u

M dx + k(y );

in this integration, y is to be regarded as a constant, and k(y) plays the role of a “constant”
of integration. To determine k(y), we derive du/dy from (6), use (4b) to get dk/dy, and
integrate dk/dy to get k. (See Example 1, below.)

Formula (6) was obtained from (4a). Instead of (4a) we may equally well use (4b).
Then, instead of (6), we first have by integration with respect to y

(6*)

u

N dy + l(x).

To determine l(x), we derive du/dx from (6*), use (4a) to get dl/dx, and integrate. We
illustrate all this by the following typical examples.

An Exact ODE

Solve

(7) cos ( x + y) dx + (3y 2 + 2 y + cos (x + y)) dy = 0.

Solution. Step 1. Test for exactness. Our equation is of the form (1) with

M = cos (x + y),

N = 3 y 2 + 2y + cos ( x + y).

Thus

— = “sin (.v + y),
dy

dN

— = -sm (x + y).
dx

From this and (5) we see that (7) is exact.

Step 2. Implicit general solution. From (6) we obtain by integration

(8) u = I M dx + k(y) = | cos (x + y) dx + k(y ) = sin (x + y) + k(y).

To find k(y), we differentiate this formula with respect to y and use formula (4b), obtaining

du dk o

— = cos (x + y) H = N = 3 y + 2y + cos (x + y).

dy dy

Hence dk/dy = 3y 2 + 2y. By integration, k = y 3 + v 2 + c*. Inserting this result into (8) and observing (3),

u(x, y) = sin (x + y) + y 3 + y 2 = c.

SEC. 1.4 Exact ODEs. Integrating Factors

23

EXAMPLE 2

EXAMPLE 3

Step 3. Checking an implicit solution. We can check by differentiating the implicit solution u(x, y) = c
implicitly and see whether this leads to the given ODE (7):

du du 9

(9) du = — dx H dy = cos ( x + y) dx + (cos (x + y) + 3 y + 2y) dy = 0.

dx By

This completes the check.

An Initial Value Problem

Solve the initial value problem

(10) (cosy sinhx + l) dx — sin y cosh x dy = 0, y(l) = 2.

Solution. You may verify that the given ODE is exact. We find u. For a change, let us use (6*),

u = — J sin y cosh x dy + /(x) = cos y cosh x + /(x).

From this, dw/dx = cosy sinhx + dl/dx = M = cosy sinhx + 1 . Hence dl/dx = 1 . By integration, /(x) = x + c*.
This gives the general solution w(x, y) = cos y cosh x + x = c. From the initial condition, cos 2 cosh 1 + 1 =
0.358 = c. Hence the answer is cos y cosh x + x = 0.358. Figure 17 shows the particular solutions fore = 0, 0.358
(thicker curve), 1, 2, 3. Check that the answer satisfies the ODE. (Proceed as in Example 1.) Also check that the
initial condition is satisfied.

Fig. 17. Particular solutions in Example 2

WARNING! Breakdown in the Case of Nonexactness

The equation — y dx + x dy = 0 is not exact because M = — y and N = x, so that in (5), BM/By = — 1 but
BN/Bx = 1. Let us show that in such a case the present method does not work. From (6),

f Bu dk

u — \M dx + k(y) = —xy + k(y), hence — = -x H .

J dy dy

Now, Bu/By should equal N = x, by (4b). However, this is impossible because k(y) can depend only on y. Try
(6*); it will also fail. Solve the equation by another method that we have discussed.

Reduction to Exact Form. Integrating Factors

The ODE in Example 3 is —y dx + x dy = 0. It is not exact. However, if we multiply it
by 1/x 2 , we get an exact equation [check exactness by (5)!],

( 11 )

— y dx + x dy

2

X

— L dx H — dy = d ) = 0.
X X \xj

Integration of (11) then gives the general solution y/x = c = const.

24

CHAP. 1 First-Order ODEs

EXAMPLE 4

This example gives the idea. All we did was to multiply a given nonexact equation, say,
(12) P(x, y) dx + Q(x, y) dy = 0,

by a function F that, in general, will be a function of both x and y. The result was an equation

(13) FPdx + FQdy = 0

that is exact, so we can solve it as just discussed. Such a function F(x, y) is then called

an integrating factor of (12).

Integrating Factor

The integrating factor in (1 1) is F = 1/jt 2 . Hence in this case the exact equation (13) is

— y dx + xdy (y\ y

FP dx + FQ dy = — : = d — ) = 0. Solution - = c.

x z W *

These are straight lines y = cx through the origin. (Note that x = 0 is also a solution of — y dx + xdy = 0.)

It is remarkable that we can readily find other integrating factors for the equation — y dx + xdy = 0, namely,
1/y 2 , 1 /fry), and I /(x 2 + y 2 ), because

(14)

~y dx + xdy
,2

— y dx + x dy
xy

~y dx + xdy

= d arctan :

How to Find Integrating Factors

In simpler cases we may find integrating factors by inspection or perhaps after some trials,
keeping (14) in mind. In the general case, the idea is the following.

For M dx + N dy = 0 the exactness condition (5) is dM/dy = dN/dx. Hence for (13),
FP dx + FQ dy = 0, the exactness condition is

(15)

d d

— {FP) = —{FQ).
dy ax

By the product rule, with subscripts denoting partial derivatives, this gives

F y P + FP y - F X Q + FQ X .

In the general case, this would be complicated and useless. So we follow the Golden Rule:
If you cannot solve your problem, try to solve a simpler one — the result may be useful
(and may also help you later on). Hence we look for an integrating factor depending only
on one variable: fortunately, in many practical cases, there are such factors, as we shall
see. Thus, let F = Fix). Then F y = 0, and F x = F' = dF/dx, so that (15) becomes

FP y = F'Q + FQ X .

Dividing by FQ and reshuffling terms, we have

\_dF
F dx

= R,

Q\dy dxj

(16)

where

SEC. 1.4 Exact ODEs. Integrating Factors

25

THEOREM 1

THEOREM 2

EXAMPLE 5

This proves the following theorem.

Integrating Factor F(x)

If ( 12) is such that the right side R of ( 16) depends only on x, then (12) has an
integrating factor F = F(x), which is obtained by integrating (16) and taking
exponents on both sides.

(17)

Fix) = exp R(x) dx.

Similarly, if F* = F*(y), then instead of (16) we get

(18)

1 dF*

= R*,

F* dy

where

and we have the companion

Integrating Factor F*(y)

If (12) is such that the right side R* of ( 18) depends only on y, then (12) has an
integrating factor F* = F*(y), which is obtained from (18) in the form

( 19 )

F*(y) = exp

R*(y) dy.

Application of Theorems 1 and 2. Initial Value Problem

Using Theorem 1 or 2, find an integrating factor and solve the initial value problem

(20) (e x+y + ye y ) dx + (xe y - l) dy - 0, y(0) = -1

Solution. Step 1. Nonexactness. The exactness check fails:

3 P 3 x+y v v SQ s v „

— = — (e y + ye v ) = e y + e v + ye v but — = — ( xe y - 1) = e y .
dy dy dx dx

Step 2. Integrating factor. General solution. Theorem 1 fails because R [the right side of (16)] depends on
both x and y.

R

_ jf3P _ _

Q \ dy dx
Try Theorem 2. The right side of (18) is

(e x

e y + ye v - i

R*

i_t dQ _ dP\
P\dx dy J

ye

- (e y - e x

e y - ye v ) = -1.

Hence (19) gives the integrating factor F*(y) = e y . From this result and (20) you get the exact equation

( e x + y) dx + (x — e ~ y ) dy = 0.

26

CHAP. 1 First-Order ODEs

Test for exactness; you will get 1 on both sides of the exactness condition. By integration, using (4a),

u = | {e x + y) dx = e x + xy + k(y).

Differentiate this with respect to y and use (4b) to get

dk

du dk

— = x -\ = N = x — e y ,

dy dy

dy

k — e y + c*.

Hence the general solution is

u{x, y) — e x + xy + e y = c.

Setp 3 . Particular solution. The initial condition y(0) = — 1 gives u{ 0, — 1) = 1 + 0 + e = 3.72. Hence the
answer is e x + xy + e~ y = 1 + e = 3.72. Figure 18 shows several particular solutions obtained as level curves
of u(x, y) = c, obtained by a CAS, a convenient way in cases in which it is impossible or difficult to cast a
solution into explicit form. Note the curve that (nearly) satisfies the initial condition.

Step 4. Checking. Check by substitution that the answer satisfies the given equation as well as the initial
condition.

PR OBLEM SET T4

ODEs. INTEGRATING FACTORS

Test for exactness. If exact, solve. If not, use an integrating
factor as given or obtained by inspection or by the theorems
in the text. Also, if an initial condition is given, find the
corresponding particular solution.

1. 2xy dx + x 2 dy = 0

2. x 3 dx + y 3 dy = 0

3. sin x cos y dx + cos x sin y dy = 0

4 . e 3e (dr + 3rd6) = 0

5. (x z + y 2 )dx — 2xy dy = 0

6. 3(y + l) dx = 2xdy, (y + l).r -4

7. 2x tan y dx + sec 2 y dy = 0

8. e x (cos y dx — sin y dy) = 0

9. e 2x (2cosydx — sinyrfy) = 0, _v(0) = 0

10. y dx + [y + tan {x + y)] dy = 0, cos ( x + y)

11. 2 cosh x cos y dx — sinh x sin y dy

12. (2xy dx + dy)e x = 0, y(0) = 2

13. e~ v dx + e- x (-e~ v + \)dy = 0, F = e x+v

14. ( a + l)y dx + {b + \)xdy — 0, y(l) = 1,

F = x a y b

15. Exactness. Under what conditions for the constants a,
b, k, l is (ax + by) dx + ( kx + ly) dy = 0 exact? Solve
the exact ODE.

SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics

27

16. TEAM PROJECT. Solution by Several Methods.

Show this as indicated. Compare the amount of work.

(a) e v (sm\\ x dx + cosh xdy) = Oas an exact ODE
and by separation.

(b) (1 + 2x) cos ydx + dy/cosy = Oby Theorem 2
and by separation.

(c) (x 2 + y 2 ) dx — 2xy dy = 0 by Theorem 1 or 2 and
by separation with v = y/x.

(d) 3jc 2 y dx + 4x 3 dy = 0 by Theorems 1 and 2 and
by separation.

(e) Search the text and the problems for further ODEs
that can be solved by more than one of the methods
discussed so far. Make a list of these ODEs. Find

17. WRITING PROJECT. Working Backward.

Working backward from the solution to the problem
is useful in many areas. Euler, Lagrange, and other
great masters did it. To get additional insight into
the idea of integrating factors, start from a u(x, y) of
your choice, find du = 0, destroy exactness by
division by some F(x, y), and see what ODE’s
solvable by integrating factors you can get. Can you
proceed systematically, beginning with the simplest
F(x, y)?

18. CAS PROJECT. Graphing Particular Solutions.

Graph particular solutions of the following ODE,
proceeding as explained.

(21) dy — y 2 sin.r dx = 0.

(a) Show that (21) is not exact. Find an integrating
factor using either Theorem 1 or 2. Solve (21).

(b) Solve (21) by separating variables. Is this simpler
than (a)?

(c) Graph the seven particular solutions satisfying the
following initial conditions y(0) = 1, y(7r/2) = ±|,
±|, ± 1 (see figure below).

(d) Which solution of (21) do we not get in (a) or (b)?

Particular solutions in CAS Project 18

Linear ODEs. Bernoulli Equation.

Population Dynamics

Linear ODEs or ODEs that can be transformed to linear form are models of various
phenomena, for instance, in physics, biology, population dynamics, and ecology, as we
shall see. A first-order ODE is said to be linear if it can be brought into the form

(1) y + p(x)y = r(x),

by algebra, and nonlinear if it cannot be brought into this form.

The defining feature of the linear ODE (1) is that it is linear in both the unknown
function y and its derivative y = dy/dx , whereas p and r may be any given functions of
x. If in an application the independent variable is time, we write t instead of x.

If the first term is f(x)y (instead of y ), divide the equation by fix) to get the standard
form (1), with y as the first term, which is practical.

For instance, y cos x + y sin x = x is a linear ODE, and its standard form is
y + y tan x = x sec x.

The function r(x) on the right may be a force, and the solution y(x) a displacement in
a motion or an electrical current or some other physical quantity. In engineering, r(x) is
frequently called the input, and y(x) is called the output or the response to the input (and,
if given, to the initial condition).

28

CHAP. 1 First-Order ODEs

Homogeneous Linear ODE. We want to solve (1) in some interval a < x < b, call
it J, and we begin with the simpler special case that r(x) is zero for all x in J. (This is
sometimes written r(x) = 0.) Then the ODE (1) becomes

( 2 )

y + P(x)y = 0

and is called homogeneous. By separating variables and integrating we then obtain
dy

y

= —p(x)dx.

thus

In \y\ = -

p(x)dx + c*.

Taking exponents on both sides, we obtain the general solution of the homogeneous
ODE (2),

( 3 )

y(x) = ce~^ x)dx

(c = ±e c when y - ~ 0);

here we may also choose c = 0 and obtain the trivial solution y(x) = 0 for all x in that
interval.

Nonhomogeneous Linear ODE. We now solve (1) in the case that r(x) in (1) is not
everywhere zero in the interval / considered. Then the ODE (1) is called nonhomogeneous.
It turns out that in this case, (1) has a pleasant property; namely, it has an integrating factor
depending only on x. We can find this factor F(x) by Theorem 1 in the previous section
or we can proceed directly, as follows. We multiply (1) by F(x), obtaining

(1*) Fy' + pFy = rF.

The left side is the derivative (Fy)' = F'y + Fy' of the product Fy if

pFy = F'y, thus pF = F' .

By separating variables, dF/F = p dx. By integration, writing h = J p dx,

P dx,

thus

F = e h .

In |f| = h =

With this F and h' = p, Eq. (1*) becomes

h f i j f h h f i / h\f / h h

e y + h e y = e y + (e ) y = (e y) = re .

By integration,

h
e y

e r dx + c.

Dividing by e , we obtain the desired solution formula

( 4 )

y(x) = e

-h

e r dx + c ), h =

p(x) dx.

SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics

29

EXAMPLE 1

EXAMPLE 2

This reduces solving (1) to the generally simpler task of evaluating integrals. For ODEs
for which this is still difficult, you may have to use a numeric method for integrals from
Sec. 19.5 or for the ODE itself from Sec. 21.1. We mention that h has nothing to do with
h{x) in Sec. 1.1 and that the constant of integration in h does not matter; see Prob. 2.

The structure of (4) is interesting. The only quantity depending on a given initial
condition is c. Accordingly, writing (4) as a sum of two terms,

(4*)

we see the following:

y(x ) = e

-h

e h rdx + ce h .

(5) Total Output = Response to the Input r + Response to the Initial Data.

First-Order ODE, General Solution, Initial Value Problem

Solve the initial value problem

y + y tan x = sin 2x, y(0) = 1 .

Solution. Here p = tan x, r = sin 2x = 2 sin cos x, and

h = J p dx = | tan x dx = In | sec x\ .

From this we see that in (4),

e h = sec x, e~ h = cos x, e h r = (sec x){2 sin x cos x) = 2 sin x,
and the general solution of our equation is

y(jr) = cos x I 2 sin x dx + c ] = c cos x — 2 cos x.

From this and the initial condition, 1 = c • 1 — 2 • l 2 ; thus c = 3 and the solution of our initial value problem
is y = 3 cos x — 2 cos 2 x. Here 3 cos x is the response to the initial data, and —2 cos 2 jc is the response to the
input sin 2jc.

Electric Circuit

Model the RL- circuit in Fig. 19 and solve the resulting ODE for the current I(t) A (amperes), where t is
time. Assume that the circuit contains as an EMF E{t) (electromotive force) a battery of E = 48 V (volts), which
is constant, a resistor of R = 1 1 D (ohms), and an inductor of L = 0.1 H (henrys), and that the current is initially
zero.

Physical Laws. A current I in the circuit causes a voltage drop RI across the resistor (Ohm’s law) and
a voltage drop Li' = L dl/dt across the conductor, and the sum of these two voltage drops equals the EMF

(Kirchhoff’s Voltage Law, KVL).

Remark. In general, KVL states that “The voltage (the electromotive force EMF) impressed on a closed
loop is equal to the sum of the voltage drops across all the other elements of the loop.” For Kirchoff’s Current
Law (KCL) and historical information, see footnote 7 in Sec. 2.9.

Solution. According to these laws the model of the /?L-circuit is Li' + RI = E(t), in standard form

R J = m
L L '

(6)

30

CHAP. 1 First-Order ODEs

EXAMPLE 3

We can solve this linear ODE by (4) with x = t, y = 7, p = R/ L, h = ( R/L)t , obtaining the general solution

/ = + c \

By integration,

( 7 )

/ r „(R/i)' , ,,

/ = e~ (R ' Lyt - — - + c = - + ce-W*.

\L R/L

R

In our case, R/L = 11/0.1 = 110 and E(t) = 48/0.1 = 480 = const; thus,

In modeling, one often gets better insight into the nature of a solution (and smaller roundoff errors) by inserting
given numeric data only near the end. Here, the general solution (7) shows that the current approaches the limit
E/R = 48/11 faster the larger R/L is, in our case, R/L = 11/0.1 = 110, and the approach is very fast, from
below if 1(0) < 48/ 1 1 or from above if 1(0) > 48/ 1 1 . If 7(0) = 48/ 1 1 , the solution is constant (48/1 1 A). See
Fig. 19.

The initial value 7(0) = 0 gives 7(0) = E/R + c = 0, c = —E/R and the particular solution
(8) 7 = -(1 - e _(R/Mt ), thus 7 = jj( 1 - e -110t ).

R = 11 (1

0.01 0.02 0.03 0.04

I

0.05

Current I{t )

t

Fig. 19. RL-circuit

Hormone Level

Assume that the level of a certain hormone in the blood of a patient varies with time. Suppose that the time rate
of change is the difference between a sinusoidal input of a 24-hour period from the thyroid gland and a continuous
removal rate proportional to the level present. Set up a model for the hormone level in the blood and find its
general solution. Find the particular solution satisfying a suitable initial condition.

Solution. Step 1. Setting up a model. Let y{t) be the hormone level at time t. Then the removal rate is Ky(t).
The input rate is A + B cos cot, where a) = 2tt/24 = 77 / 12 and A is the average input rate; here A B to make
the input rate nonnegative. The constants A, B, K can be determined from measurements. Hence the model is the
linear ODE

y it) = In — Out = A + B cos cot — Ky(t), thus y' + Ky = A + B cos cot.

The initial condition for a particular solution y part is y pa rt(0) = y 0 with t = 0 suitably chosen, for example,
6:00 A.M.

Step 2. General solution. In (4) we have p = K = const, h — Kt, and r = A + B cos cot. Hence (4) gives the
general solution (evaluate f e Kt cos cot dt by integration by parts)

SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics

31

y(t) = e

A + B cos cot )dt + ce

A

K

K cos ojt + oj sin cot

+ ce

A

K

K 2 + (77/ nf

TTt TT . 777

K cos 1 sin —

12 12 12

The last term decreases to 0 as t increases, practically after a short time and regardless of c (that is, of the initial
condition). The other part of y(t) is called the steady-state solution because it consists of constant and periodic
terms. The entire solution is called the transient-state solution because it models the transition from rest to the
steady state. These terms are used quite generally for physical and other systems whose behavior depends on time.

Step 3. Particular solution. Setting t = 0 in y(t) and choosing y 0 = 0, we have

y(0) =

B

K + c = 0,

thus

K K 2 + (tt/12) 2 77
Inserting this result into y(t), we obtain the particular solution

_ _A _

KB

K K 2 + (tt/12) 2

ypart^O "L

B

K K 2 + (tt/12) 2

TTt 7T

TTt

K cos — H — — sin — — 1 — I 1-

12 12

12

KB

K K 2 + (tt/12) :

with the steady-state part as before. To plot _y paI1 we must specify values for the constants, say, A = B = 1
and K = 0.05. Figure 20 shows this solution. Notice that the transition period is relatively short (although
K is small), and the curve soon looks sinusoidal; this is the response to the input A + Bcos(jjTTf) =

1 + COS (pj TTt).

y

25

20

15

10

5 ■

100

200

Fig. 20. Particular solution in Example 3

Reduction to Linear Form. Bernoulli Equation

Numerous applications can be modeled by ODEs that are nonlinear but can be transformed
to linear ODEs. One of the most useful ones of these is the Bernoulli equation 7

( 9 )

y + P(x)y = g(x)y a

(, a any real number).

7 JAKOB BERNOULLI (1654—1705), Swiss mathematician, professor at Basel, also known for his contribution
to elasticity theory and mathematical probability. The method for solving Bernoulli’s equation was discovered by
Leibniz in 1696. Jakob Bernoulli’s students included his nephew NIKLAUS BERNOULLI (1687-1759), who
contributed to probability theory and infinite series, and his youngest brother JOHANN BERNOULLI (1667-1748),
who had profound influence on the development of calculus, became Jakob’s successor at Basel, and had among
his students GABRIEL CRAMER (see Sec. 7.7) and LEONHARD EULER (see Sec. 2.5). His son DANIEL
BERNOULLI (1700-1782) is known for his basic work in fluid flow and the kinetic theory of gases.

32

CHAP. 1 First-Order ODEs

EXAMPLE 4

If a = 0 or a = 1, Equation (9) is linear. Otherwise it is nonlinear. Then we set

u(x) = [j(.r)] 1- “.

We differentiate this and substitute y from (9), obtaining

u = (1 - a)y~ a y' = (1 - a)y~ a (gy a - py).

Simplification gives

«'=(!- a)(g ~

where v 1_a = u on the right, so that we get the linear ODE
(10) u + (1 — a)pu = (1 — a)g.

For further ODEs reducible to linear form, see lnce’s classic [All] listed in App. 1. See
also Team Project 30 in Problem Set 1.5.

Logistic Equation

Solve the following Bernoulli equation, known as the logistic equation (or Verhulst equation 8 ):

(11) y = Ay - By 2

Solution. Write (11) in the form (9), that is,

y' - Ay = ~ By 2

to see that a = 2, so that u = )> 1-a = y -1 . Differentiate this u and substitute y' from (11),
u = ~y~ 2 y' = -y~\Ay - By 2 ) = B - Ay' 1 .

The last term is — Ay -1 = —An. Hence we have obtained the linear ODE

u + Au = B.

The general solution is [by (4)]

u = ce~ At + B/A.

Since u = 1 /y, this gives the general solution of (11),

( 12 )

B/A

Directly from (11) we see that y = 0 (y(t) = 0 for all t) is also a solution.

(Fig. 21)

8 PIERRE-FRAN£OIS VERHULST, Belgian statistician, who introduced Eq. (8) as a model for human
population growth in 1838.

SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics

33

EXAMPLE 5

Fig. 21. Logistic population model. Curves (9) in Example 4 with A/B = 4

Population Dynamics

The logistic equation (11) plays an important role in population dynamics, a field
that models the evolution of populations of plants, animals, or humans over time t.
If B = 0, then (11) is y' = dy/dt = Ay. In this case its solution (12) is y = (1 /c)e At
and gives exponential growth, as for a small population in a large country (the
United States in early times!). This is called Malthus’s law. (See also Example 3 in
Sec. 1.1.)

The term —By 2 in (11) is a “braking term” that prevents the population from growing
without bound. Indeed, if we write y = Ay [ 1 — (B/A)y], we see that if y < A/B. then
y > 0, so that an initially small population keeps growing as long as y < A/B. But if
y > A/B, then y < 0 and the population is decreasing as long as y > A/B. The limit
is the same in both cases, namely, A/B. See Fig. 21.

We see that in the logistic equation (11) the independent variable t does not occur
explicitly. An ODE y = fit, y) in which t does not occur explicitly is of the form

(13) y' = fiy)

and is called an autonomous ODE. Thus the logistic equation (11) is autonomous.

Equation (13) has constant solutions, called equilibrium solutions or equilibrium
points. These are determined by the zeros of fiy), because fiy) = 0 gives y = 0 by
(13); hence y = const. These zeros are known as critical points of (13). An
equilibrium solution is called stable if solutions close to it for some t remain close
to it for all further t. It is called unstable if solutions initially close to it do not remain
close to it as t increases. For instance, y = 0 in Fig. 21 is an unstable equilibrium
solution, and y = 4 is a stable one. Note that (11) has the critical points y = 0 and
y = A/B.

Stable and Unstable Equilibrium Solutions. “Phase Line Plot”

The ODE y' = (y — l)(y — 2) has the stable equilibrium solution yx = 1 and the unstable y 2 — 2, as the direction
field in Fig. 22 suggests. The values y x and y 2 are the zeros of the parabola /(y) = (y — l)(y — 2) in the figure.
Now, since the ODE is autonomous, we can “condense” the direction field to a “phase line plot” giving yx and
y 2 , and the direction (upward or downward) of the arrows in the field, and thus giving information about the
stability or instability of the equilibrium solutions.

34

CHAP. 1 First-Order ODEs

yU )

1 1 1 1 1 1 / ts.a

t t t t t tt t t t
//////////
/////////s'

1

tt 1 1 1 1 1 1 1 1
/ / / / / / / / / /
//////////

*2;

'//// ’ 0 : 5 '

//////////

//////////

4 / / / // / / / / / / 4 / / / /

//////////

//////////

/////////:

l

t-

^2

■ yi

2.0 -

1.5 -

0

1.0 -

0.5 -

Vi

u 2 1 | 3* 2

0.5 1.0 1.5 2.0 2.5 3.0 *

(a) (6) (c)

Fig. 22. Example 5. (A) Direction field. (B) “Phase line”. (C) Parabola f(y)

A few further population models will be discussed in the problem set. For some more
details of population dynamics, see C. W. Clark. Mathematical Bioeconomics : The
Mathematics of Conservation 3rd ed. Hoboken, NJ, Wiley, 2010.

Further applications of linear ODEs follow in the next section.

-In x

= l/x(not —x) and

1. CAUTION! Show that e

e — ln(secx) = cosx

2. Integration constant. Give a reason why in (4) you may
choose the constant of integration in fp dx to be zero.

3-13

GENERAL SOLUTION. INITIAL VALUE

PROBLEMS

Find the general solution. If an initial condition is given,
find also the corresponding particular solution and graph or
sketch it. (Show the details of your work.)

3. y' — y = 5.2

4. y = 2y — 4x

5. y' + ky = e~ kx

6. y + 2y = 4 cos 2x, y{\l r) = 3

7. xy = 2v + x 3 e x

8. y' + ytanx = e“ 001l cosr, y(0) = 0

9. y + ysinx = e cosx , y(0) = -2.5

10. y' cos x + (3y — l)sec.r = 0, yi^Tr) — 4/3

11 . y = (y - 2) cot x

12. xy + 4y = 8x 4 , y( I ) = 2

13. y = 6 (y — 2.5)tanh 1.5x

14. CAS EXPERIMENT, (a) Solve the ODE y - y/x =
—x _1 cos (l/x).Find an initial condition for which the
arbitrary constant becomes zero. Graph the resulting
particular solution, experimenting to obtain a good
figure near x = 0.

(b) Generalizing (a) from n = 1 to arbitrary n, solve the
ODE y — ny/x = — x” -2 cos (1/x). Find an initial
condition as in (a) and experiment with the graph.

15-20

GENERAL PROPERTIES OF LINEAR ODEs

These properties are of practical and theoretical importance
because they enable us to obtain new solutions from given
ones. Thus in modeling, whenever possible, we prefer linear
ODEs over nonlinear ones, which have no similar properties.

Show that nonhomogeneous linear ODEs (1) and homo-
geneous linear ODEs (2) have the following properties.
Illustrate each property by a calculation for two or three
equations of your choice. Give proofs.

15. The sum + y 2 of two solutions yq and y 2 of the
homogeneous equation (2) is a solution of (2), and so is
a scalar multiple ay 1 for any constant a. These properties
are not true for (1)!

SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics

35

16. y = 0 (that is, y(x) = 0 for all x, also written y(x) = 0)
is a solution of (2) [not of (1) if r(x) A 0!], called the

trivial solution.

17. The sum of a solution of (1) and a solution of (2) is a
solution of (1).

18. The difference of two solutions of (1) is a solution of (2).

19. If yi is a solution of (1), what can you say about cyi?

20. If yi and y 2 are solutions of y[ + py i = tq and
y 2 + py 2 — r 2 , respectively (with the same p\), what
can you say about the sum yi + v 2 ?

21. Variation of parameter. Another method of obtaining
(4) results from the following idea. Write (3) as cy*,
where y* is the exponential function, which is a solution
of the homogeneous linear ODE y *' + py* = 0.
Replace the arbitrary constant c in (3) with a function
u to be determined so that the resulting function y — uy*
is a solution of the nonhomogeneous linear ODE
y + py — r.

22-28

NONLINEAR ODEs

Using a method of this section or separating variables, find
the general solution. If an initial condition is given, find
also the particular solution and sketch or graph it.

22. / + y = y 2 , y(0) = -§

23. y + xy = xy -1 , v(0) = 3

24. y + y = —x/y

25. y = 3.2y - 10y 2

26. y = (tan y)/{x - 1), y(0) = j7T

27. y = \/{6e y - 2x)

28. 2xyy' + (x — l)y 2 = x 2 e x (Sety 2 = z)

29. REPORT PROJECT. Transformation of ODEs.

We have transformed ODEs to separable form, to exact
form, and to linear form. The purpose of such
transformations is an extension of solution methods to
larger classes of ODEs. Describe the key idea of each
of these transformations and give three typical exam-
ples of your choice for each transformation. Show each
step (not just the transformed ODE).

30. TEAM PROJECT. Riccati Equation. Clairaut
Equation. Singular Solution.

A Riccati equation is of the form

(14) y + p(x)y = g(x)y 2 + h{x).

A Clairaut equation is of the form

(15) y — xy' + g(y').

(a) Apply the transformation y = Y + l/u to the
Riccati equation (14), where Tis a solution of (14), and
obtain for u the linear ODE u + (2 Yg — p)u = —g.
Explain the effect of the transformation by writing it
as y = Y + v, v = l/u.

(b) Show that y = Y = x is a solution of the ODE
y — (2x 3 + 1 ) y = — x 2 y 2 — x 4 — x + 1 and solve this
Riccati equation, showing the details.

(c) Solve the Clairaut equation y' 2 — xy' + y = 0 as
follows. Differentiate it with respect to x, obtaining
y"(2 y - x) = 0. Then solve (A) y" = 0 and (B)
2y — x = 0 separately and substitute the two solutions
(a) and (b) of (A) and (B ) into the given ODE. Thus
obtain (a) a general solution (straight lines) and (b) a
parabola for which those lines (a) are tangents (Fig. 6
in Prob. Set 1.1); so (b) is the envelope of (a). Such a
solution (b) that cannot be obtained from a general
solution is called a singular solution.

(d) Show that the Clairaut equation (15) has as
solutions a family of straight lines y = cx + g{c) and
a singular solution determined by g'(s) — —x, where
s = y , that forms the envelope of that family.

31-40

MODELING. FURTHER APPLICATIONS

31. Newton’s law of cooling. If the temperature of a cake
is 300°F when it leaves the oven and is 200°F ten
minutes later, when will it be practically equal to the
room temperature of 60°F, say, when will it be 61°F?

32. Heating and cooling of a building. Heating and
cooling of a building can be modeled by the ODE

T' = k,(T - T a ) + k 2 (T - TJ + P,

where T = T(t) is the temperature in the building at
time ?, T a the outside temperature, T w the temperature
wanted in the building, and P the rate of increase of T
due to machines and people in the building, and kj and
k 2 are (negative) constants. Solve this ODE, assuming
P = const, T w = const, and T a varying sinusoidally
over 24 hours, say, T a = A — Ccos(27r/24)t.Discuss
the effect of each term of the equation on the solution.

33. Drug injection. Find and solve the model for drug
injection into the bloodstream if, beginning at t = 0, a
constant amount A g/min is injected and the drug is
simultaneously removed at a rate proportional to the
amount of the drug present at time t.

34. Epidemics. A model for the spread of contagious
diseases is obtained by assuming that the rate of spread
is proportional to the number of contacts between
infected and noninfected persons, who are assumed to
move freely among each other. Set up the model. Find
the equilibrium solutions and indicate their stability or
instability. Solve the ODE. Find the limit of the
proportion of infected persons as t — * 00 and explain
what it means.

35. Lake Erie. Lake Erie has a water volume of about
450 km 3 and a flow rate (in and out) of about 175 km 2

36

CHAP. 1 First-Order ODEs

per year. If at some instant the lake has pollution
concentration p = 0.04%, how long, approximately,
will it take to decrease it to p/2, assuming that the
inflow is much cleaner, say, it has pollution
concentration p/4, and the mixture is uniform (an
assumption that is only imperfectly true)? First guess.

36. Harvesting renewable resources. Fishing. Suppose
that the population y(t) of a certain kind of fish is given
by the logistic equation (11), and fish are caught at a
rate Hy proportional to y. Solve this so-called Schaefer
model. Find the equilibrium solutions Vi and y 2 (> 0)
when H < A. The expression Y = Hy z is called
the equilibrium harvest or sustainable yield corre-
sponding to H. Why?

37. Harvesting. In Prob. 36 find and graph the solution
satisfying y(0) = 2 when (for simplicity) A = B = 1
and H = 0.2. What is the limit? What does it mean?
What if there were no fishing?

38. Intermittent harvesting. In Prob. 36 assume that you
fish for 3 years, then fishing is banned for the next
3 years. Thereafter you start again. And so on. This is
called intermittent harvesting. Describe qualitatively
how the population will develop if intermitting is
continued periodically. Find and graph the solution for
the first 9 years, assuming that A = B = 1, H = 0.2,
and y(0) = 2.

Fig. 23. Fish population in Problem 38

39. Extinction vs. unlimited growth. If in a population
y (t) the death rate is proportional to the population, and
the birth rate is proportional to the chance encounters
of meeting mates for reproduction, what will the model
be? Without solving, find out what will eventually
happen to a small initial population. To a large one.
Then solve the model.

40. Air circulation. In a room containing 20,000 ft 3 of air,
600 ft 3 of fresh air flows in per minute, and the mixture
(made practically uniform by circulating fans) is
exhausted at a rate of 600 cubic feet per minute (cfm).
What is the amount of fresh air y(t) at any time if
y(0) = 0? After what time will 90% of the air be fresh?

1.6 Orthogonal Trajectories. Optional

An important type of problem in physics or geometry is to find a family of curves that
intersects a given family of curves at right angles. The new curves are called orthogonal
trajectories of the given curves (and conversely). Examples are curves of equal
temperature (isotherms) and curves of heat flow, curves of equal altitude (contour lines)
on a map and curves of steepest descent on that map, curves of equal potential
(equipotential curves, curves of equal voltage — the ellipses in Fig. 24) and curves of
electric force (the parabolas in Fig. 24).

Here the angle of intersection between two curves is defined to be the angle between
the tangents of the curves at the intersection point. Orthogonal is another word for
perpendicular.

In many cases orthogonal trajectories can be found using ODEs. In general, if we
consider G(x, y, c) = 0 to be a given family of curves in the xy-plane, then each value of
c gives a particular curve. Since c is one parameter, such a family is called a one-
parameter family of curves.

In detail, let us explain this method by a family of ellipses

(1)

(c > 0)

SEC. 1.6 Orthogonal Trajectories. Optional

37

and illustrated in Fig. 24. We assume that this family of ellipses represents electric
equipotential curves between the two black ellipses (equipotential surfaces between two
elliptic cylinders in space, of which Fig. 24 shows a cross-section). We seek the
orthogonal trajectories, the curves of electric force. Equation (1 ) is a one-parameter family
with parameter c. Each value of c (> 0) corresponds to one of these ellipses.

Step 1. Find an ODE for which the given family is a general solution. Of course, this
ODE must no longer contain the parameter c. Differentiating (1), we have x + 2yy = 0.
Hence the ODE of the given curves is

(2) y =f(x,y) = - J *-.

2 y

Fig. 24. Electrostatic field between two ellipses (elliptic cylinders in space):
Elliptic equipotential curves (equipotential surfaces) and orthogonal
trajectories (parabolas)

Step 2. Find an ODE for the orthogonal trajectories y = y(x). This ODE is

( 3 )

r

l

fix, y)

= +

2 y

x

with the same /as in (2). Why? Well, a given curve passing through a point (x 0 > y 0 ) has
slope /(x 0 , To) at that point, by (2). The trajectory through (x 0 , y 0 ) has slope — 1 //(jc 0 , To)
by (3). The product of these slopes is —1, as we see. From calculus it is known that this
is the condition for orthogonality (perpendicularity) of two straight lines (the tangents at
(x 0 , .Vo)), hence of the curve and its orthogonal trajectory at (x 0 , y 0 ).

Step 3. Solve (3) by separating variables, integrating, and taking exponents:

dy -Ci be .|~i - . ~ *2

— = 2 — , In \y | = 2 In x + c, y = c x .

y x

This is the family of orthogonal trajectories, the quadratic parabolas along which electrons
or other charged particles (of very small mass) would move in the electric field between
the black ellipses (elliptic cylinders).

38

CHAP. 1 First-Order ODEs

FTOB =EEra=SFF=F=g

1-3

FAMILIES OF CURVES

Represent the given family of curves in the form
G(x, y; c) = 0 and sketch some of the curves.

1. All ellipses with foci —3 and 3 on the x-axis.

2. All circles with centers on the cubic parabola y = x 3
and passing through the origin (0, 0).

3. The catenaries obtained by translating the catenary
y — cosh x in the direction of the straight line y = x.

ORTHOGONAL TRAJECTORIES (OTs)

Sketch or graph some of the given curves. Guess what their
OTs may look like. Find these OTs.

4. y = x 2 + c 5. y = cx

6. xy = c 7. y = c/x 2

8. y = Vx + c 9. y = ce~ x

10 . x 2 + (y - cf = c 2

APPLICATIONS, EXTENSIONS

11. Electric field. Let the electric equipotential lines

(curves of constant potential) between two concentric
cylinders with the z-axis in space be given by
u(x, y) = x 2 + y 2 = c (these are circular cylinders in
the xyz-space). Using the method in the text, find their
orthogonal trajectories (the curves of electric force).

12. Electric field. The lines of electric force of two opposite
charges of the same strength at ( — 1,0) and (1,0) are
the circles through ( — 1, 0)and (1,0). Show that these
circles are given by x 2 + (y — c) 2 = 1 + c 2 . Show
that the equipotential lines (which are orthogonal
trajectories of those circles) are the circles given by
(x + c*) 2 + y 2 = c* 2 — 1 (dashed in Fig. 25).

Fig. 25. Electric field in Problem 12

13. Temperature field. Let the isotherms (curves of
constant temperature) in a body in the upper half-plane
y > 0 be given by 4x 2 + 9y 2 = c. Find the ortho-
gonal trajectories (the curves along which heat will
flow in regions filled with heat-conducting material and
free of heat sources or heat sinks).

14. Conic sections. Find the conditions under which
the orthogonal trajectories of families of ellipses
x 2 /a 2 + y 2 /b 2 = c are again conic sections. Illustrate
CAS. What happens if a -> 0? If b -> 0?

15. Cauchy-Riemann equations. Show that for a family
u(x, y) = c = const the orthogonal trajectories v(x, y) =
c* = const can be obtained from the following
Cauchy-Riemann equations (which are basic in
complex analysis in Chap. 13) and use them to find the
orthogonal trajectories of e x sin y = const. (Here, sub-
scripts denote partial derivatives.)

U x Vy , Uy V x

16. Congruent OTs. Ify r =/(x) with /independent of y,
show that the curves of the corresponding family are
congruent, and so are their OTs.

Existence and Uniqueness of Solutions
for Initial Value Problems

The initial value problem

|y'| + W=0, y(0) = 1

has no solution because y = 0 (that is, y(x ) = 0 for all x) is the only solution of the ODE.
The initial value problem

y' = 2x,

y( 0) = l

SEC. 1.7 Existence and Uniqueness of Solutions

39

has precisely one solution, namely, y = x 2 + 1. The initial value problem

xy' = y - l, y(0) = l

has infinitely many solutions, namely, y = 1 + cx, where c is an arbitrary constant because
y(0) = 1 for all c.

From these examples we see that an initial value problem

(1) y = f(x, y), y{xf) = y 0

may have no solution, precisely one solution, or more than one solution. This fact leads
to the following two fundamental questions.

Problem of Existence

Under what conditions does an initial value problem of the form (1) have at least
one solution ( hence one or several solutions)?

Problem of Uniqueness

Under what conditions does that problem have at most one solution ( hence excluding
the case that is has more than one solution)?

Theorems that state such conditions are called existence theorems and uniqueness
theorems, respectively.

Of course, for our simple examples, we need no theorems because we can solve these
examples by inspection; however, for complicated ODEs such theorems may be of
considerable practical importance. Even when you are sure that your physical or other
system behaves uniquely, occasionally your model may be oversimplified and may not
give a faithful picture of reality.

THEOREM 1

Existence Theorem

Let the right side f(x, v) of the ODE in the initial value problem

(1) y' =f(x,y), y(x 0 ) = yo

be continuous at all points ( x , y) in some rectangle

R: \x - x 0 \ < a, |y - y 0 | < b (Fig. 26)

and bounded in R; that is, there is a number K such that

(2) | f(x, y)\ = K for all (x, v) in R.

Then the initial value problem (1) has at least one solution y(x). This solution exists
at least for all x in the subinterval \x — jtol < a of the interval \x — xol <
here, a is the smaller of the two numbers a and b/K.

40

CHAP. 1 First-Order ODEs

y

R

?

1

1

1

Fig. 26. Rectangle R in the existence and uniqueness theorems

( Example of Boundedness. The function /(x, y) = x 2 + y 2 is bounded (with K = 2) in the
square |x| < 1 , |_y | < 1. The function f(x, y) = tan (x + y) is not bounded for
|x + y| < 77/2. Explain!)

THEOREM 2

Uniqueness Theorem

Let f and its partial derivative f y = df/dy be continuous for all (x, y) in the rectangle
R (Fig. 26) and bounded, say,

(3) (a) |/(x, y) | ^ K, (b) \f y (x,y)\ ^ M for all (x, y) in R.

Then the initial value problem (1) has at most one solution y(x). Thus, by Theorem 1,
the problem has precisely one solution. This solution exists at least for all x in that
subinterval \x — XqI < a.

Understanding These Theorems

These two theorems take care of almost all practical cases. Theorem 1 says that if f(x, y)
is continuous in some region in the xv-plane containing the point (xo, VoX then the initial
value problem ( 1 ) has at least one solution.

Theorem 2 says that if, moreover, the partial derivative df/ dy of / with respect to y
exists and is continuous in that region, then (1) can have at most one solution; hence, by
Theorem 1, it has precisely one solution.

Read again what you have just read — these are entirely new ideas in our discussion.

Proofs of these theorems are beyond the level of this book (see Ref. [All] in App. 1);
however, the following remarks and examples may help you to a good understanding of
the theorems.

Since y' = f(x,y), the condition (2) implies that \y' Si K\ that is, the slope of any
solution curve y(x) in R is at least —K and at most K. Hence a solution curve that passes
through the point (xo, yo) must lie in the colored region in Fig. 27 bounded by the lines
1 1 and 1 2 whose slopes are —K and K, respectively. Depending on the form of R, two
different cases may arise. In the first case, shown in Fig. 27a, we have b/K a and
therefore a = a in the existence theorem, which then asserts that the solution exists for all
x between xo — a and xo + a. In the second case, shown in Fig. 27b, we have b/K < a.
Therefore, a = b/K < a, and all we can conclude from the theorems is that the solution

SEC. 1.7 Existence and Uniqueness of Solutions

41

EXAMPLE 1

exists for all x between x 0 — b/K and xo + b/K. For larger or smaller x’s the solution
curve may leave the rectangle R, and since nothing is assumed about / outside R , nothing
can be concluded about the solution for those larger or amaller x’s; that is, for such x’s
the solution may or may not exist — we don’t know.

(a) ( 6 )

Fig. 27. The condition (2) of the existence theorem, (a) First case, (b) Second case

Let us illustrate our discussion with a simple example. We shall see that our choice of
a rectangle R with a large base (a long x-interval) will lead to the case in Fig. 27b.

Choice of a Rectangle

Consider the initial value problem

y' = 1 + v 2 , y(0) = 0

and take the rectangle R: \x\ < 5, |y| < 3. Then a = 5, b = 3. and

\f(x,y)\

V

dy

= |l + y 2 | £ K = 10,
= 2|y| £ M = 6,

a

< a.

Indeed, the solution of the problem is y = tan* (see Sec. 1.3, Example 1). This solution is discontinuous at
±77/2, and there is no continuous solution valid in the entire interval |*| < 5 from which we started.

The conditions in the two theorems are sufficient conditions rather than necessary ones,
and can be lessened. In particular, by the mean value theorem of differential calculus we
have

fix, y 2 ) - f(x, >’ 1 )

(*2 “ Ti)

dy

y

y

where (x, yi) and (x, y 2 ) are assumed to be in R, and y is a suitable value between yi
and y 2 - From this and (3b) it follows that

(4) I/O, y 2 ) ~ fix, Vi)| S M\y 2 — ^il -

``` 